help-17

[sly mountain]

wut

[finite mist]

Then u made a mistake along the way u get that number

[kind light]

bro you cant just send us some random number and ask for help without context

[vast shale]

i thought maybe you guys know how to break that point thing into simpler terms so that i can solve this

[sly mountain]

faktz

[finite mist]

Solve it? The calc has already solved it for u

Why torturing yourself....

[vast shale]

ok ! maybe the book answer could be wrong too

i have verified my sum its coming to that point and then by using calculator the final answer doesn't match the option

.close

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[steep radish]

How to do this?

[steep radish]

I keep getting alot of cases to find

[kind light]

i dont think bruteforcing is that crazy?

just check for 2xx, 3xx, 4xx, ect

[steep radish]

I tried and it kept on going tho

Should i try again?

Okay yeah got it

.close

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[neat rose]

Let $a,b,c,d > 0 : a+b+c+d = 3$; Prove : $$\sum_{cyc} (abc)^3 \leq 1$$

[twin meteor]

Goëtia

[dense cedar]

This was an application of the multinomial theorem though !

[neat rose]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[dense cedar]

What does cyc under the summation sign means ?

[hybrid flicker]

cycle

so a,b,c then b,c,d then c,d,a...

[dense cedar]

Okay !

[neat rose]

if someone has a basic idea , ping me, i dont want to overkill it with something that the basic highschooler doesnt know

[shut canyon]

have you tried cubing the equation

[neat rose]

u try it

[shut canyon]

😭

i give up 🥲

[dense cedar]

Maybe you would like to use langrange multipliers

Or ik the max value is at a=b=c=d

Or
(abcd)³(1/a³+1/b³+1/c³+1/d³)
So (abcd)^(1/4) ≤ 3/4
So (abcd)³ ≤ (3/4)^12
Idk what follows after

[paper depot]

wait hold on couldn't we AM-GM this somehow

[shut wave]

ye but wed need d tho

why d excluded?

[dense cedar]

/ see this

[shut wave]

oh

[neat rose]

as i said no overkill method to be used

[paper depot]

(abc)^3 <= (a+b+c)^3/27 and likewise under cyclical shifts

[shut wave]

hmm

[paper depot]

is that cooking?

[neat rose]

no

[paper depot]

sad! many such cases

[neat rose]

exactly

there is a basic method im sure it exists

[neat rose]

also ur wrong here

u might have meant abc instead of (abc)^3

[paper depot]

oh yes i did mean that my bad

was trying to transform the ineq in my head and messed it up

ok that's definitely doing us no good then

[neat rose]

i bet my right arm, the trick is to use jensen's inequality on f(x) = x^3

but not sure how to use it without including that the sum is maximized at a=b=c=d=3/4

[shut wave]

when does the cycle end?

[neat rose]

Aight i think i might have found a clue

lets use what Ann-san used

abc <= ((a+b+c)/3)^3

abc <= ((3 -d)/3)^3

thus the cyclic sum of (abc)^3 <= ((3-d)/3)^9 + ... + ((3 - a)/3)^9

aight now consider g(x) = ((3-x)/3)^9

this thing is clearly convex for x in [0, 3]

then by jensen ineq

(g(a) + g(b)+g(c)+g(d))/4 >= g(a+b+c+d/4)

= g(3/4)

[hybrid flicker]

so you're on the wrong side of Jensen

[neat rose]

what u talkin bout?

[hybrid flicker]

you wanted to upper bound (g(a) + g(b)+g(c)+g(d))/4

[neat rose]

na chill

we'll get there soon

[neat rose]

here is an exercise for you, prove that g(x) gets its max at x = 0

yeah i got it thx everyone

.close

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[hybrid flicker]

not sure how you got it but alr

[hybrid flicker]

• 1. obvious
• 2. not here to solve your exercises

[neat rose]

(g(a)+g(b) + g(c)+g(d))/4 <= 1

u asked the obvious, i gave u an exercise to help u see the obvious as u fail to see it

[hybrid flicker]

and yet we're asking you to bound g(a) + g(b) + g(c) + g(d)

not (g(a) + g(b) + g(c) + g(d))/4

[neat rose]

whats the monotony of g(x) over [0,3]

[hybrid flicker]

so congratulations, your upper bound is 4

[hybrid flicker]

yes, decreasing, so g(x) <= g(0), so your upper bound is 4g(0) = 4

not a great upper bound imo

Please stop acting condescending if you want to receive help

[neat rose]

not condescending, i said i got it, u kept pushin it, not sure if u got offended

[hybrid flicker]

I pushed it because it appeared you didn't solve the original question (which indeed you didn't) and I wanted to help you realize it. You give me "exercises" to "see the obvious as I fail to see it"

sounds pretty condescending

[neat rose]

good luck

[hybrid flicker]

no, I wish you good luck, it is you who has an unresolved question

[neat rose]

resolved it, good luck the channel was marked closed

[hybrid flicker]

xddddd

[bitter pilot]

xddd

[vocal sleet]

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[sudden tundra]

hello: given X and N, what is Y where X = (Y * (Y+1) * (Y+2) * (Y+3) * ... * (Y+N)) / (N+1)!

[hybrid flicker]

so X = (Y+N) choose (N+1)

[sudden tundra]

X = (Y * (Y+1) * (Y+2) * (Y+3) * ... * (Y+N)) / (N+1)!

[hybrid flicker]

yes

X and N are natural integers?

[sudden tundra]

yes

[hybrid flicker]

or only N?

[sudden tundra]

i'm not sure how to formualte the equation given a non-integer for N

[hybrid flicker]

well if you want to extend to any X,Y,N

$X = \frac{(Y+N)!}{(N+1)!(Y-1)!}$

[twin meteor]

rafilou is not not born in 2003

[hybrid flicker]

so extend using Gamma function, which generalizes factorials

$X = \frac{\Gamma(Y+N+1)}{\Gamma(N+2)\Gamma(Y)}$

[twin meteor]

rafilou is not not born in 2003

[sudden tundra]

X = (Y * (Y+1) * (Y+2) * (Y+3) * ... * (Y+N)) / (N * (N - 1) * (N - 2) * (N - 3) * .. * (N - N +1)

what is that R symbol?

[hybrid flicker]

Gamma function

Gamma(m+1) = m!

[sudden tundra]

ok

[hybrid flicker]

in any case you have this

for natural integers X,N,Y

not sure how to extract a formula for Y though

[hard atlas]

for which values of X and N are you asked that question?

you could always go the bruteforce way, multiply out the polynomial and use rrt. if you know that Y is an integer then you are guaranteed to find it. it will just be painful

depending on size it will certainly be faster to just write down pascals triangle

you can also do things with the prime factorization. you have to allocate the primes in a way so that you get a product of consecutive integers

depending on size you can also just approximate X approx Y^N/(N+1)! and take the Nth root of X(N+1)!. that will give you an approximate Y and then you can look at the nearby integers and see if they work

[vocal sleet]

@sudden tundra Has your question been resolved?

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[waxen hawk]

Given there are 5 numbers which have a sum of 45, prove that one of these number must be equivalent or bigger than 9.

[waxen hawk]

.close

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[waxen hawk]

.reopen

[vocal sleet]

✅

[waxen hawk]

Question:

Given there are 5 numbers which have a sum of 45, prove that one of these number must be equivalent or bigger than 9.

This one is tougher than I expected

[hybrid flicker]

!status

[vocal sleet]

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

[waxen hawk]

I wqs thinking that I might be able to prove it if I introduce the geometric meaning of a weighted mean

[hybrid flicker]

oh uh no need to go complicated

[hybrid flicker]

let's analyse this statement you're tasked to prove

Here's what I see (and tell me if you agree with me or not)

[waxen hawk]

Yes, that would take too much effort to explain how it works

[hybrid flicker]

I see three parts in that statement

"There are 5 numbers"

"which have a sum of 45"

prove
"one of these numbers is equal or bigger than 9"

right now nothing fancy, right?

[waxen hawk]

Sure

Just words of statements

[hybrid flicker]

This is the premise, nothing out of the ordinary

so say we call them a,b,c,d,e

we don't really care about their name either

[waxen hawk]

I did, when I was trying manipulating stuff to an answer

[hybrid flicker]

Now, what kind of statement is this?

Are we meant to prove it

or...

[waxen hawk]

You mean do the other way around, proof by contradiction

Is it how it is called

[hybrid flicker]

I'm not going there yet, but you're smelling what I'm cooking

[hybrid flicker]

This part of the statement, do we have to prove it?

What function does it serve

how is it called

[waxen hawk]

Is it the if not B then not A

[hybrid flicker]

ok you're going 1000% faster than I was anticipating

[waxen hawk]

If the sum isn’t 45 then the premise wouldn’t hold

[hybrid flicker]

no that's not it

[waxen hawk]

Ohhh my bad

[hybrid flicker]

let's just go through with what I wanted you to find

[hybrid flicker]

This part of the statement is called a hypothesis/supposition

[hybrid flicker]

and this is the result

so we're in a "prove A => B" situation

there are generally three ways to prove such a statement

you have the direct proof:

"Suppose A.
....
then B"

and then you have 2 additional kinds of proof

the contrapositive proof

which relies on the fact that "A => B" is the same as "not B => not A"

(recall the ever so great 'if I poop then I flush the toilet')

[waxen hawk]

I see

[hybrid flicker]

so the proof would go
"Suppose the result is false...

Then the hypothesis is false"

And final kind of proof that we won't use here

Proof by absurd

"Suppose the hypothesis is true but the result is false.
...
Wait, that's impossible!!!"

[waxen hawk]

If there is no number that bigger or equal to 9, or in other words, a,b,c,d, and e are all smaller than 9.

Then a+b+c+d+e would be smaller than 45.

Hence, if the sum of these five constants is 45, then one of them must be equivalent or bigger than 9.

[hybrid flicker]

by contrapositive

yep, you got it

[waxen hawk]

It is easier than I expected

[hybrid flicker]

so when you're tasked to prove statements that look like A => B

before adventuring in a direct proof

try playing around with what the contrapositive is like, or what a proof by absurd would look like

and see if one of those is more obvious than the others

[hybrid flicker]

cause we can agree that this statement isn't obvious at first

but if I write it in its contrapositive form "given there are 5 numbers who are all smaller than 9, prove that they have a sum not equal to 45"

definitely looks easier

[waxen hawk]

I never thought contraposition to be that useful and commonplace.

It is fabulous, thanks for providing such a wonderful method of proving.

[vocal sleet]

@waxen hawk Has your question been resolved?

[vocal sleet]

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[errant flint]

heyy

[errant flint]

I need help with a division

that I cant solve it in the calculator

but its very bigs numbers

[reef agate]

Can you show the division problem

[errant flint]

and Im kinda struggling

okk

[reef agate]

Do you know how to do long division?

[errant flint]

hm I think no

what u mean by long division?

[paper depot]

https://es.wikipedia.org/wiki/División_larga

[errant flint]

yea I dont know how to do that

wait

I know how to do that

I tried to do that

[paper depot]

can you show an attempt

[errant flint]

yea

[paper depot]

ok let's see it then

[errant flint]

there

[paper depot]

uh oh

why are you bringing three digits down

you're supposed to only bring one down at a time 😭

[errant flint]

really?

[paper depot]

yes

also might help to write down the first 9 multiples of 197 just to not have to recalculate them every time

[errant flint]

ok

so

I bring only the 3 down?

[paper depot]

no

you start with the 310 as you did

[errant flint]

ok

[paper depot]

but in the next step you dont touch the 5 and 6 yet

and just bring 1 down so you have 1131

and all subsequent steps you bring down 1 digit at a time

[errant flint]

yea I did that

I put a line

to seperate

[paper depot]

oh ok

[errant flint]

between the 1 and 5

[paper depot]

well then you want to know how many times 197 goes into 1131

[errant flint]

yea

but I wanna know if theres any trick

[paper depot]

if you are doing it by hand, it's helpful to note that 197 is approximately 200

[errant flint]

to make it easier

[errant flint]

okk

and that 1131 its close to 1130?

[paper depot]

1131 is between 1000 and 1200

[errant flint]

oh ok

[paper depot]

so 200 goes into it 5 times but not 6

same for 197 but double-check it

(if you end up with more than 197 as remainder, you can always just subtract another 197)

well, this or the more general trick is to write out 197*1, 197*2, 197*3, ..., 197*9 and compare your current remainder against all of those

[errant flint]

wait

[errant flint]

I understood this part

but when I have to subtract

I dont know what is the number I use

like

Idk how much is 197*5

[paper depot]

do you know how to multiply

or at least how to add

[errant flint]

I forgot Im used to use a calculator

but I cant use it

[paper depot]

well you have to relearn how to add and multiply

without these you have no hope of doing division by hand

[errant flint]

okk

Ill be back

[errant flint]

Im back

got it

[paper depot]

you forgot manual addition and multiplication, but now you are back and have relearned it in all of 2 minutes?

[errant flint]

yea

[paper depot]

either you are a lightning fast learner or something's suspicious here.

but ok, can you work out say 197*7 on paper for me and show me the result?

as in all your work not only the answer

[errant flint]

yea

Im sending u the picture

@paper depot

[paper depot]

,rcw

[paper depot]

careful with that 9, it's easy to misread as a 4

but ok seems good enough

[errant flint]

okk

[paper depot]

should i repeat my instructions for the division problem?

[errant flint]

nah

thx for your help

[paper depot]

ok

[errant flint]

wait

Im doin now 320/45

and I did

that 50 can enter 6 in 300

[paper depot]

320/45?

a different question?

[errant flint]

32/45

[paper depot]

...

[errant flint]

but I added the zero

wait

[paper depot]

you are confusing me

[errant flint]

Im gonna send a photo

[paper depot]

you are doing a different question than the one we started with, yes or no?

[errant flint]

yea I know Im confused too ahahhahah

[errant flint]

its the same exercise

[paper depot]

huh??

[paper depot]

then why are you dividing anything by 45 when your divisor was 197??

[errant flint]

wait

this is a different question but they both are from the same exercise

[paper depot]

ok

is it 32 or 320 then

[errant flint]

32

[paper depot]

so it should be 32.0 in your working

not 320

[errant flint]

yea wait

I know

[paper depot]

anyway,

[errant flint]

but

[paper depot]

i did say that estimation with rounding can sometimes mean you are off by 1

[paper depot]

here

[errant flint]

as 32 is smaller than 45 I added a zero after 32 and a zero before 45

[paper depot]

i see your issue: 320 - 270 = 50 and that can fit another 45

[errant flint]

okk

[errant flint]

ok

well now Im good

thx

[vocal sleet]

@errant flint Has your question been resolved?

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[white spoke]

Hello sorry to bother you, but i have got a question for 2.3 i have not done a question like that in a very long time could you please help?

[white spoke]

I have got a test comming up on monday aswell so need to revise some stuff for maths

[flat whale]

,rotate

[flat whale]

find all the factors of 36 and 63

if they don't share any factors, then the HCF is just the product 36 * 63

[white spoke]

So with factors do you mean like one factor of 36 is 18?

[flat whale]

right

,w factors of 36

[twin meteor]

Results provided by WolframAlpha

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

Requested by rie.mann

[flat whale]

prime factors would make it easier to compare, but not necessary

[white spoke]

Oh ok i get it now and so what ever is the highest factor that is in both the numbers is the correct answer?

[flat whale]

https://www.cuemath.com/numbers/hcf-highest-common-factor/

[white spoke]

Ok thanks

Thats all my questions!

I'll close the chat now

Cya

.close

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[bleak pasture]

Man idk whats wrong with my awnsers?

[paper depot]

also you have a doubled plus sign in there

that and also you should show your work for all of these answers

it's not really possible to diagnose your errors just from the wrong answers here

[bleak pasture]

I fixed that

[bleak pasture]

I got the slope by doing the slope equation y1-y-2/x1-x2

Then plugged xandy

To get h

B*

and i got -10p+1930

[paper depot]

ok theres multiple issues with that

one is that you're missing the parentheses

[bleak pasture]

?

[paper depot]

two is that you're throwing away the variable names and dangerously untethering yourself from the problem's variables (and have to then get the translation back from x and y to p and q correct)

three is that this planning is all well and good but we need to see the execution as well.

show dont tell, etc.

an arithmetic error could befall absolutely ANYBODY and we need to make sure this didnt happen either

[bleak pasture]

Let me redo it because i didnt it on a throwaway paper

[paper depot]

ok

[bleak pasture]

Yeah i fricked up on the equation thank you

I did a number incorrectly

.close

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[vocal sleet]

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[jovial copper]

help

close

[twin horizon]

Type .close

@jovial copper

[jovial copper]

.close

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[severe stone]

So I dont i am allowed to pick 1 quarter of this square and then multiply my answer by 4 right?

[flat whale]

not unless your function is the same on all 4 quadrants

which i doubt

[severe stone]

ye exactly that is what i was thinking about

so what then tho

[flat whale]

do 4 double integrals

[severe stone]

is that the fastest way?

[flat whale]

find some symmetry and maybe do 2instead

[severe stone]

aaah in the function right?

do I use even and oneven stuff then?

[flat whale]

or describe D with 2 regions

so many ways to do it just try any of them

[severe stone]

alright!

so i figured f(x,-y) = f(x,y) does this mean that the function is symmetrical in y=0 so the x-asis?

[civic otter]

I'm not so sure about that

Because you have a y³

[severe stone]

oh shit

i thought y²

i wrote it down wrong

[civic otter]

O would integrate it either with the horizontal "wires" (blue) or with the vertical "wires" (green)

[severe stone]

aaha ye

and split it just in 2

-1 to 0

and 0 to 1

for either x or y

[civic otter]

Exactly

[severe stone]

thank you

[tight delta]

can i skip the nomials section including bi,tri,polynomials? how important is it for future maths?

[severe stone]

so R1= { (x,y) element R² | -1<=x<=0 , -x-1<=y<=x+1 }?

[severe stone]

!occupied

[vocal sleet]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tight delta]

come again?

[severe stone]

I am already using this channel for help so could u please create one yourself?

[tight delta]

so sorry...will do

[severe stone]

no worries :)

[severe stone]

is it true that this one takes really long? maybe was easier to do it horizontal?

but that will screw with the sin(x)

[civic otter]

Uh yeah I didn't even think about that sine

I believe horizontally is much better then

[vocal sleet]

@severe stone Has your question been resolved?

[severe stone]

well the sine will be anoying if i do it horizontally right?

but now i am doing it vertically but makes a huge function for the outside integration

[severe stone]

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[frosty thorn]

,,3sec(x)-4cos(x)=0

[twin meteor]

☠ cj Σ

[frosty thorn]

only in [0,2pi)

this is my work

[bitter pilot]

cos(x) = 0 must not be

because you are dividing by 0 else

[frosty thorn]

oh i c

so is it just pi/3 and 5pi/3

[bitter pilot]

they are wrong

you should have 3sec²(x)-4=0

you ignored the 3 when you factored

[frosty thorn]

i c

i missed that

ty

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[dire meteor]

.open

[eager oxide]

where does this 8 come from?

[eager oxide]

nvmmm

[sly sierra]

look at the integration bounds

[eager oxide]

i didnt realize they changed

oops

[sly sierra]

yea they're using that the integrand is even

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[rich sinew]

I just need some clarification on a simultanious equaitons question

[rich sinew]

it wants me to solve this for a does not equal 1/2 and a does equal 1/2

this was the first thing i did

if a does not equal 1/2, y is equal to 0 and after that getting x and z is pretty easy

but i dont see how a = 1/2 has a different solution

the answers use a parameter for y and then solves it from there

[vocal sleet]

@rich sinew Has your question been resolved?

[sinful heron]

if a = 1/2, then y can be anything

if a is not equal to 1/2, y has to be 0, otherwise your last equation does not hold, you understand?

[rich sinew]

oh ok thanks

i thought i had a way of obtaining y=0 without using a but i just realised it had an error lol

i understand alot better now

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[weary forge]

How do I Graph/write Quadratics in standard form?

[weary forge]

I'm on this question

[mental cradle]

i would draw each different part, and then add em up
first draw 2x^2, then -4x, then -6

thats a practical way, dk what you need tho

[weary forge]

I need to graph the equation in standard form ig

Parabola

[mental cradle]

well then thats good

[weary forge]

Alright, I did it

[mental cradle]

for each vertical line, add the value of each 'part', at the center it should be -6

i guess you can assume each square means only 1+-1

+-1

[weary forge]

???

My teacher had been explaining this, but I was absent cause I was sick, so I have no idea what to do

Apparently the x-coordinate of the vertex is - b/2a

Which in this case should be 4/2(2)

Which is 4/4, which simplified would just be 1

But plugging 1 into the equation doesn't make sense when I do the equation...

[snow dune]

what happens when you plug 1 into the equation?

[weary forge]

Well it works out fine, but the video I watched said I had to plug something else in after 1, but the number that I plug in after leads to -144, and I don't think that's right...

[snow dune]

if you need to find the vertex, just plugging in 1 is fine.

[weary forge]

I get -6

[snow dune]

should be .8

-8

[weary forge]

I tried that

But that would be 2(-8)^2

[snow dune]

thats not what i mean sorry

[weary forge]

Oh

[snow dune]

if you plug in 1, the end result should be -8

[weary forge]

How?

[snow dune]

2*(1)^2-4(1)-6

[weary forge]

2(1)^2, doesn't that equal 4?

[snow dune]

2(1)^2 is the same as 2 * 1 * 1

[weary forge]

I though it would be the same as 2^2

[snow dune]

Im not quite sure how u get that

maybe ur multiplying before doing the exponent

you have to do exponents before multiplication

[weary forge]

Ohhhh ok

Even then, aren't I supposed to plug in -10 and/or -6 after?

[weary forge]

I know that, but 2 away from -8 are -10 and -6

[wraith python]

(x,y) = (1, -8)

That gives you a coordinate pair whose point you can plot on your graph.

That point also happens to be the vertex of the parabola.

[weary forge]

I was thinking -8 was the X, I'm so stupid

Thank you

Yknow I think I get it now

[wraith python]

🙂

[weary forge]

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[lofty sparrow]

We started doing volume by slicing and I can't make sense of it I am genuinely lost on where to even start

[lofty sparrow]

this is the problem I'm currently stuck on but if someone could help me by giving me an example problem and helping guide me through it so I understand what I am doing that would be a huge help

[trim fern]

so it's like a skewed semicone ish shape

oh wait the questions are simpler than i thought

so like

(for part 1) the diameter of the semicircle increase as X increases

being 0 at x=2 and 2 at x=0

simple linear function of 2-x

and then just add the other part of the formula to transform the diameter into the semicircle's area

like

since a(d) = pi*d^2/8

then A(x) = pi/8 * (2-x)^2

[lofty sparrow]

Well in class he usually tells us to find what we are respecting to like y or x and we find the area of a slice then use an integral to find the whole area its just when Im looking at it I got no clue what im respecting which is the first step to solving this I think

and I think its respect to x since its perpendicular to the x

[trim fern]

i thought of a solution w/o calculus

so

imagine if the half cone was unskewed

wait a second i forgot the cone volume formula LMAO brb

[lofty sparrow]

is there a good visual for that cause idk what unskewed is

[trim fern]

so like
if the base triangle had vertices (0,0) (0,2) (1,0)

then it'd look like /\ while having the same area

now if we extend the resulting half cone to a full cone

we get one with diameter 2 and height 2

(=> radius 1)

its volume is then 1/3 * h * pi * r^2
= 1/3 * 2 * pi * 1
= 2pi/3

we halve it to getr the half cone's volume

so pi/3

and the original shape should also have the volume pi/3

[lofty sparrow]

bro lowkey I se the vision

cause there was also a another similar method from class ;llike rotating the shape around an axis to get the full shape and such and then finding the area based off that

[trim fern]

thought i'd double check w/ the integral but then i remembered that idk how to integrate 💀

[lofty sparrow]

😭

its alg

Ill try it rq and see what I come up with

but you are right the area is pi/3 which is nuts cause I was trying to do it the way they taught in class which was super complicated and lowkey got me mind boggled and you solved w/o it got like wwwwoooowww

[trim fern]

yea wolframalpha confirms that the integral is pi/3

[lofty sparrow]

and lowkey changed my perspective on the problem

yeah Im putting togther the inegral rn

see if I can come up with pi/3

yeah my brains fried I cant figure out how to put the integral togtyher its like I almost got it but then my answer came out to 8pi/3

im missing smthing

[vocal sleet]

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[lofty sparrow]

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[near kiln]

Hi, how can I start doing this?

[near kiln]

It suggests doing a substitution but I don’t see how that can help

[bitter pilot]

have u tried it

[grim lotus]

you won't see how it helps unless u try it

[vocal sleet]

@near kiln Has your question been resolved?

[near kiln]

If we differentiate u with respect to x, we get something similar to what is inside the integrand, but I still dont see what I have to do

[leaden ingot]

I would not follow the suggestion.

Start by multiplying and dividing by x².

Then use the substitution u = x+1/x

[vocal sleet]

@near kiln Has your question been resolved?

[bitter pilot]

,w Integrate[(x^2-1)/(x^2+1) * 1/sqrt(1+x^4), x]

[twin meteor]

Results provided by WolframAlpha

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

Requested by adoniszgks

[bitter pilot]

what a waste of time

[leaden ingot]

$\frac{x^2-1}{x^2+1}\cdot \frac{1}{\sqrt{1+x^4}}=\frac{1-\frac{1}{x^2}}{x+\frac{1}{x}}\cdot \frac{1}{\sqrt{x^{2}+\frac{1}{x^2}}}$

[twin meteor]

TargetVN

[leaden ingot]

i am very disappointed that you didnt even respond

$u=x+\frac{1}{x} \text{ then }du=\left( 1-\frac{1}{x^2} \right),dx$

[twin meteor]

TargetVN

[leaden ingot]

i hope it is clear enough

if you dont understand or just dont like this apporach, at least respond instead of completely ignoring me and marking this problem as "unsolved"

[bitter pilot]

hi to bye 💀

[vocal sleet]

@near kiln Has your question been resolved?

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[sharp bay]

hey, would it be possible for someone to help me figure out how to make smooth connections between the overlapping points in my graph, they have to be algebraically proven with derivatives
https://www.desmos.com/calculator/kyn7xrgm7s

[wraith python]

You know the derivatives need to be equal at x = a. Use that to determine the coefficient of a functions antiderivative.

[sharp bay]

sorry what

[wraith python]

So you have one of these equations above already, f(x).

That is the curve that goes through the stem.

To the right of that curve, you have the equation 0.6x^2 + 8.12.

Change that to b x^2 + c.

You can solve for an arbitrary x value so that the derivative is equal at x. Once you find that x value, you can solve for c so that the curves intersect.

[sharp bay]

I did some calculations for 1st and 3rd function on desmos and it didn’t work

[wraith python]

One moment ...

[sharp bay]

im doing the calculations for this right now

[vocal sleet]

@sharp bay Has your question been resolved?

[sharp bay]

okay but what do these functions mean- i understand the steps but i dont get what the final functions are. 😭 🙏

tqsm btw

[wraith python]

Have you learned about initial value problems?

[sharp bay]

no i havent

[wraith python]

You have learned about derivatives though, correct?

[sharp bay]

yeah i have but some parts are still iffy

[wraith python]

Do you understand what an anti-derivative is?

[sharp bay]

i havcent learnt about that yet

i know how to derive but this whole concept of smooth function is confusing

[wraith python]

Ok. Well an antiderivative is the opposite of a derivative.

The caveat is that an anti-derivative adds a constant c because when you take a derivative of a function, any constants disappear so you have to add them back when you calculate an anti-derivative.

If you reverse the process of of finding the derivative of 2x, you add 1 to the exponent of x and then divide by that new exponent to reduce the coefficient.

[sharp bay]

i haven't learnt integrals yet either

so far I've only learnt differentiation

[wraith python]

Here, I am adding 1 to the exponent of x instead of subtracting.

I am also dividing by that same value because previously you brought it down.

Does that make sense?

Have you learned how to do derivatives without using limits?

[sharp bay]

i haven't learnt about limits yet i think

[wraith python]

Are you in Calculus or Pre-calculus?

I assume at least Calculus since you mentioned smooth curves and differentiable.

[sharp bay]

yes

i just finished pre calc

[wraith python]

Weird. This is a bit of an advanced project if you don't know about limits, smooth curves, and such.

[sharp bay]

im pretty sure ive done limits

but my teacher didn't go in depth

this is all ive done

[wraith python]

This is related to anti-derivatives.

You basically find x^n by adding **1 **to (n-1) and divide by n - 1 + 1 or just n.

[olive stag]

an antiderivative is when u ask, if i know the derivative of f(x) was blah, what must the original function have been?

but u will learn it when u learn it

but taht is the basic concept

[wraith python]

This project may be too advanced for them. It requires an understanding of anti-derivatives, initial value problems, smoothness, and differentiability.

[sharp bay]

the way all my friends are doing this is that there creating a function and messing with the points until both of the functions derivatives match

[wraith python]

Is this an extra credit problem?

[olive stag]

i had a similar project in numerical analysis

[sharp bay]

no this is 20% of my final grade

[olive stag]

jinkies

[wraith python]

I mean, it's a trivial exercise if you know what you're doing.

Shoudn't take more than 30 minutes.

[olive stag]

wait so what exactly are u trying to do?

alter the parameters so u have a nice apple?

[wraith python]

They are trying to create smooth curves for the apple above.

[olive stag]

if she alters the bounds of each function and makes the connection points have the same slope it will work

[sharp bay]

yea I guess i need to have at least 3 smooth functions in my design for a logo which in my case is an apple

[wraith python]

But she is having trouble with making the points of intersections differentiable.

[sharp bay]

originally i kept messing with points until i got them to connect but the gradient of the function were different even though they were connected

[wraith python]

I wrote this for her which shows the step-by-step process, but she doesn't understand what much of it means.

[sharp bay]

i understand most of it

[wraith python]

Which parts confused you?

[sharp bay]

i dont get the last part

so when you assume x=1

i get it till there

but everything after that is a bit confusing

[wraith python]

I used x = 1 again because that is what I used to solve for the value of the derivaties.

The part after that relates to anti-derivatives.

There is a constant c that you must always account for when solving anti-derivatives, because constants are zeroed out when you differentiate it.

That is why I added a + c to g(x).

[remote kraken]

Hey im doing the same assignment as her and here's an example of how the teacher wants us to do it (although im missing alot of steps and might not be accurate)

[olive stag]

woah

[wraith python]

Yeah, that's doing the exact same thing.

[olive stag]

what a weird problem to give to calc 1 students

[remote kraken]

Yea i didn't know how to explain it to her

[remote kraken]

Its for maths advanced (australia)

[wraith python]

You definitely need to have learned about initial value problems and anti-derivatives to know how to do this. It doesn't require guesswork.

[wraith python]

Do you understand the steps in the image I posted?

[remote kraken]

Yea i sorta get it

[wraith python]

Do both of you understand how to find an anti-derivative?

[sharp bay]

i don't

[remote kraken]

Not rly

Yo if its not too crazy cld we ask if u can help on call? 😭

She lowk crashing out

[wraith python]

Do both of you want to learn?

[olive stag]

the problem of trying to connect two functions smoothly is one I saw first in a numerical analysis class, a bit more advanced than what ive seen in calc 1 where u r just introduced to derivatives

But basically the concept on how to smoothly connect two functions f(x) and g(x)
is to ask

1. at what x value do I wana connect the two points? Let's call this x = a
2. condition: make f(a) = g(a) so its connected
3. condition: make f'(a) = g'(a) so its smooth

[remote kraken]

Yes pls

[wraith python]

Let me write something out.

That's a quick and dirty explanation.

The latter half is what is being done here.

[olive stag]

i tried solving what the shared slope would be for the two lines that make the top of the apple
one is
f(x)=ax^2+b
the other
g(x)=Acos(Bx+C)+D
and jinkies, idk how to solve for the parameters w/o num analysis stuff

[wraith python]

This.

[olive stag]

i think its just being sleep deprived but i cannot read that for the life of me rn

oh wait

[sharp bay]

i did the calcs and put it into desmos but its not working 😭

[olive stag]

.

[wraith python]

f'(x) = g'(x) to solve for differentiability and f(x)=g(x) for continuity.

[olive stag]

lol

dont forget the period for seriousness

[sharp bay]

This is the work I did for some of the functions before

[wraith python]

You know what. Let's try something easier first. What you are doing for now is a little too complicated.

[sharp bay]

do u think u can vc and explain if its possible 🙏

im so lost

[wraith python]

video chatting won't necessarily make this easier.

Let's say you have this function, f(x) = x^2, that has the interval (-inf, 1].

[sharp bay]

yea

[olive stag]

ohhh ur fixing the parameters of one and seeing what the paremeters of the next function should be

[wraith python]

And you want to extend that curve to the right so that it is smooth, meaning differentiable at x=1.

[olive stag]

dadoi i wasnt assuming that, i was so confused as to how to get the parameters

[wraith python]

Now, you can determine the derivative of f(x) at x = 1 rather easily. It's just f'(x) = 2x for x = 1.

[olive stag]

it makes sense cause u gotta start SOMEWHERE

[wraith python]

f'( 1 ) = 2(1) = 2.

Does this make sense so far?

[sharp bay]

yea i get it

[wraith python]

Let's make the simplest function we can that has a derivative of 2. The simplest example is a linear function, g(x) = 2x.

Does this make sense so far?

[sharp bay]

yes

[wraith python]

@remote kraken You still with us?

Now you can obviously see that the two equations do not intersect at x = 1.

[remote kraken]

Yea

[sharp bay]

yes

[wraith python]

We need to vertically shift g(x) so that it does intersect with f(x) at x = 1.

That vertical shift is the c constant that I wrote and your instructions also wrote.

[sharp bay]

oh ok

[wraith python]

Same idea is being used for both examples.

We adjust one of the equations vertically so that they intersect at x = 1.

f(x) = g(x) + c

x^2 = 2x + c

We solve for x = 1 because that is where we want the two equations to intersect.

f(1) = g(1) + c

(1)^2 = 2(1) + c

1 = 2 + c

c = -1

[sharp bay]

ok yea i get that

[wraith python]

Now the two equations intersect at x = 1, continuous, and they have the same derivative for x = 1 at x = 1, differentiable.

[sharp bay]

ok yea i get it

would this work with functions im using

[wraith python]

Yes, look back at my image that I posted.

The only difference is that instead of using a distinct function, I used the general form of a function.

[sharp bay]

ok yea

i think i get what you mean

[wraith python]

One moment, let me use graph what I mean.

So on the top of the apple, you have these two functions and you want them to intersect at some point as well as be differentiable at that same point.

You need to generalize one of the two functions. The easiest one to generalize is g(x).

And determine at what value of x you want it to intersect.

Here, the graph is broken up into units of 0.2.

It would make sense to have it intersect at maybe x = 0.4.

So you use g(x) = ax^2 + b and f(x) as written above. We use g(x) in a general form because polynomial functions are more easier to work with than trigonometric functions.

[sharp bay]

ok ill try do that

[wraith python]

You will need to leave x in unsimplified form because if you simplify it, the approximation will result in rounding errors.

[sharp bay]

Idk if I’m on the right track

[wraith python]

One moment.

Solve for the derivative being equal first.

And then solve for f(x) = g(x) + c

You need to solve for differentiability first and then solve for continuity second.

And remember that you need the general form of either function.

This is the general form of either function. As you can see, the trigonometric function has a lot more unknown variables that you would need to solve for which is why we choose the general form of g(x).

[wraith python]

Having any luck?

Honestly, I would try avoiding using trigonometric functions if possible. They make this a lot more difficult.

[sharp bay]

im thinking that the 2 parabolas at the bottom and top i should scratch put from my design and just try connect the side function

im trying to solve what you told me right now and i think i am getting it

https://www.desmos.com/calculator/pcmsmjhyfh

[wraith python]

Cool. Well I'm off for a bit, if you have any further questions regarding this problem, feel free to ping me.

[sharp bay]

ok thank you for your help i really appreciated it

[sly mountain]

ou

[vocal sleet]

@sharp bay Has your question been resolved?

[vocal sleet]

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[north totem]

i want to find the ? in this riddle

[upper shell]

Is there no other information given?

[atomic jasper]

what is the 20 referring to?

[upper shell]

area of ABX

its blue

[atomic jasper]

QX=BX and PX=AX

[upper shell]

how?

[atomic jasper]

from similiar triangles

[upper shell]

it isnt given that AB is parallel to DC

[atomic jasper]

actually

thats not true

lol

im stupid

[upper shell]

thats why i asked this

[mild flower]

it's not given, but it's also not contradicted. therefore you can assume it's true

[upper shell]

oh yeah thats smart

[mild flower]

you can also assume ABPQ is a rectangle if it makes it easier

or even a square if you want

[upper shell]

but then we cant prove that the area of the orange part is whatever we get

[mild flower]

why not

[upper shell]

cus of our assumption

we need to be sure of all cases

[mild flower]

if the problem has an answer, then you'll find it

if the problem does not have an answer then it's not a valid problem

[upper shell]

i mean yeah

but were cheesing it through and through

ok ima try it with the assumption

[mild flower]

what you can do at that point is unconstrain it a little bit (make it a rectangle vs a square) and see if that changes

[upper shell]

yeah im getting the answer with the parallel assumption

[mild flower]

@north totem ^

[north totem]

yeah?

[mild flower]

yeah what do you think about the discussion so far on your problem

[upper shell]

ok first of all is there any information you havent given us

[north totem]

let me see

ab is parallel to cd cp=pq=qd=ab and the cut at x the S of BXA is 20 what is the S of BCPX

[upper shell]

great

ab is parallel is the important thing

!status

[vocal sleet]

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

[north totem]

2

i wanted to find the area of abcp and minus 20

OH WAIT

[upper shell]

im sure there are many ways to do it

yeah?

[north totem]

i think px is = to ax

[upper shell]

yeah

[north totem]

bc ax/xp = dq/pq

which is 1

[upper shell]

dp/pq?

[north totem]

sorry dq and pq

[upper shell]

but you gotta prove that

ok how about this

[north totem]

thats is a rule

[upper shell]

can you prove some relation between triangle ABX and XQP

[north totem]

bc qx is parallel to AD

[upper shell]

is that given in the question statement?

[north totem]

no but it is given that AB is parallel to DC and that means taht abqd is a parallelogram

[upper shell]

yeah you're right

thats one way

so with this information

that px=ax\

and we also know that ab = qp

can you prove congruence between triangle ABX and XQP

[north totem]

yeah

[upper shell]

youre right (we also show that <PAB = <APQ cus AB and QP are parallel so we get SAS congruency)

by this congruency we get that their areas are equal

so are of XQP = 20

and also that BX = XQ

now see triangle BQC

we know that area of XQP = 20

and that XP joins the midpoints of the two sides

from this can you get the orange area

[north totem]

bc its a midsegment?

[upper shell]

kinda yeah

think what the orange area is in triangle BQC

its area of BQC - area of XQP

we know the area of XQP

can you get the area of BQC from the area of XQP

[north totem]

oh wait i think that 2px is bc

[upper shell]

ye

try to prove similarity of BQC and XQP

[north totem]

qp/qx=qc/qb

so they are simmilar

which mean that their areai

is a/b^2 Xs

a/b being their similarrity diffrentce

[upper shell]

go on

[north totem]

'יןבי ןד 2 אם 1

which is 2 to 1

so that means that the area of the big one is 4 times biggers

so i think its 80

[upper shell]

exactly

now get the orange area

[north totem]

dude how didnt i get it on the test

its 80-20

so its 60

[upper shell]

superb

this was on the test?

[north totem]

yeah

i am so stupid

[upper shell]

nah nah nah

we all learn

[north totem]

its for the national olimpic in math

i think i might have not passed

i am at teh national team for coding tho

[upper shell]

nice i love coding

[north totem]

can i give u unother one that was on there?

[upper shell]

sure

[north totem]

first just so you know a calc isnt allowed

[upper shell]

ok

[north totem]

in an progression that always goes up they are those numbers 7,8 ,A,B,70,80 how many ways can a b not make an Arithmetic progression that is 4 long

so A B cannot be 9 10 50 60 and whatever it is fir 8 and 70

so there are 3 pairs that are illigel so i wanted to say the amout of pairs minus 3

but i didnht have time to count the amount of pairs

[upper shell]

yeah counting prolly isnt correct

ok if we first choose A

we can choose from 9 to 68

thats 60 options

[north totem]

true

wait why 60

and not 59

[upper shell]

and if we choose 9,50 or ?

[upper shell]

if we can choose from 1 to 5 tahts not 5-1 = 4 options

[north totem]

yeah right

[upper shell]

wait is that any natural solution to the 8 and 70 progression

if the common diff is d

then A = 8 + d

[north totem]

39 i thinkg

[upper shell]

B = 8 + 2d

70 = 8 + 3d

=> d = 62/3

=20.6

theres no whole number solution so we dont gotta care about it

[north totem]

wdym?

oh yeah

[upper shell]

theres is no two natural numbers A and B such that 8,A,B,70 is ap

[north totem]

yeah

[upper shell]

anyways if we choose A to be 9,50 B cant be 10, 60

[north totem]

so its the amount of leagel pairs minus 2

[upper shell]

youre right

so to calculate the amount of pairs

for each A we choose we can get upto 69-A B's

[north totem]

no

bc a is less the b

so if we chose a 50 b must be more the 50

ans also less then 70

[upper shell]

im talking about the number of B's we can choose

so if we take any A

B has to fall in the range of A+1 to 69

[north totem]

yeah

[upper shell]

which is equal to 69 - (A+1) + 1 = 69-A

(made a mistake earlier)

[north totem]

what those this mean

[upper shell]

so the number of legal pairs should be $\sum_{i=9}^{i<=68} 69-i$

[twin meteor]

CherryMan

[north totem]

yeah but there is no calc

[upper shell]

we can just use formulas

this is basically 60 + 59 + 58 + 57 + ... + 1

= 60(61)/2 = 1830

1830 -2 = 1828

[upper shell]

did you get this part atleast

[north totem]

yeah

[upper shell]

great then we just expand it

[upper shell]

to get this

[north totem]

i think this + 60 tho

[upper shell]

?

[north totem]

this is the amount of b right?

ooh nvm

[upper shell]

this is the number of total pairs

[north totem]

ok litsen what i can say here t osolve this easely

[north totem]

.

[upper shell]

wait can i explain my reasoning

[north totem]

sure

[upper shell]

if we choose A=9, we can choose 69-9 = 60 Bs
if we choose A=10, we can choose 69-10 = 59 Bs
if we choose A=11, we can choose 69-11 = 58 Bs
...
if we choose A=68, we can choose 69-68 = 1 B

[north totem]

yeah

[upper shell]

then the some of the total number of B's for every A we choose

is the sum of all legal pairs

= 60 + 59 + ... + 1

[north totem]

-2

[upper shell]

yaeh

[north totem]

yeah so to calc this without using a calc

i think what u do is

i can say that the first and the last 1 is 61 and the second and the second to last is also 61 so i can do the 61 times the amount of pairs devided by 2 i think

[upper shell]

you just invented the sum of natural numbers formul;a yourself

congratulations!!!

[north totem]

but there might be an edge case for add amount of numbers

[upper shell]

this is literally the proof for the fact that 1+2+3+...+n = n(n+1)/2

[north totem]

ok i think if there is an odd amount u take out the middle one do what i said then add it aswell

[upper shell]

ittl give the same formula

ittl still be n(n+1)/2

[north totem]

does my thing work for odd amount of numbers too?

[upper shell]

yeah

universal

[north totem]

how??

[upper shell]

if n is odd

then if we first take the sum of n-1 numbers

= (n-1)(n)/2

and add n

= (n-1)(n)/2 + n = n(n+1)/2

[north totem]

oh yeah

[upper shell]

so calculate the number of total pairs

with the formula

[north totem]

dude i didnt have time ti calculate it in the test