help-10

[timid silo]

Need help, stuck with Q2c

[pliant grove]

hi

emmm..

[timid silo]

Hi :))

How do I show that F(k+1) is 2^k

[pliant grove]

mmm.. the alternatives

?

[timid silo]

Strong Induction, the TA is gonna be picky if I pick another method

[pliant grove]

ahh ok ok

mmm.... wait

I don't know I'm not sure I think 25

;<

need help

please

i don't know

please help me

[timid silo]

You know normal Induction?

Induction with inequalities?

[pliant grove]

No, I don't know, only the exercise says to calculate, that's as far as I know.

[pliant grove]

oh yes yes

[timid silo]

Strong induction applies more to recursive functions

This is example with strong Induction

[pliant grove]

ahh, yes I know but it doesn't work for me or maybe there is another solution please help

[pliant grove]

oh it's spectacular

[timid silo]

I have trouble with this form of induction too 😓😓

[pliant grove]

but i need a your help please

[timid silo]

I need help too 🤣

[pliant grove]

xd

[timid silo]

We in this together 😂

[pliant grove]

yess

[timid silo]

So... how gonna get help 🤔

[pliant grove]

I think the same

[timid silo]

.close

[obtuse pebble]

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[late roost]

Hi so im learning how to find the perticular solution to this equation, The lecture is saying that because G(x)=x^2, y has to be a polinomial of degree 2. Can anyone explain why that is?

[obtuse pebble]

@late roost Has your question been resolved?

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[patent marten]

Ok so I'm confused as to what the answer for this problem is. My professor said it is zero, but I said DNE because it's breaking one of the limit laws.

[obtuse pebble]

@patent marten Has your question been resolved?

[patent marten]

<@&286206848099549185>

[vernal spear]

i think u should it to different server and see whether someone may help

[patent marten]

Yeah this will be the 3rd time sadly just going to give up on this one

.close

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[timid silo]

low key confusing 9th grade math

im kinda bad sry just learned this todayu

nvm solved on my own lmao

[obtuse pebble]

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[umbral hornet]

why must n be a positive integer?

[zenith raft]

'must' is the wrong word here

it's not saying it must be an integer and otherwise the equality doesn't hold

[umbral hornet]

I'm not quite understanding what you mean

[violet sentinel]

you used the word "must" but it's not saying "must".

[umbral hornet]

okay that i understand but

why does it say n is a positive integer?

[zenith raft]

under the right hypotheses you can extend that to n being a real number and still have the equality

[umbral hornet]

this is my first time doing limits so sorry if i sound dumb here 🥲

[zenith raft]

but it's making it easier for you

(i assume the question is asking you to prove that)

[brave bramble]

There's cases where the equation might fail, if n is not an integer

[umbral hornet]

oh i see

may i have an example?

[brave bramble]

lim[x -> 0] √[x] is undefined

But √[lim[x -> 0] x] = 0

The problem is the discontinuity of √x, at the limit point.

[umbral hornet]

i see

thank you 👍

.close

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[obtuse pebble]

@calm axle Has your question been resolved?

[obtuse pebble]

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[gritty swallow]

hello, how would you go about part c in this question

[brave bramble]

Simplify 5 - r/10

[gritty swallow]

oh i have to use a trig identity i see

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[bright geyser]

not really sure

[tough bolt]

The part that you colored blue is the part that B-A and C-A shares, so given that B-A=C-A mean that the C-A-B region and B-A-C region are empty

Or else there will be elements that are in B or C that isn't in the blue part

[obtuse pebble]

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[grim wharf]

It would be helpful if someone can explain the concept behind this sort of problem as well it would be much appreciated

[obtuse pebble]

@grim wharf Has your question been resolved?

[grim wharf]

<@&286206848099549185>

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[narrow garnet]

how do you solve using hyperbolic functions? I tried integration by parts but i got stuck. u = tanh^-1(x), du = 1/1-x^2, dv = x and v = x^2/2. when i do the integral of v du, im not sure how to proceed

[obtuse pebble]

@narrow garnet Has your question been resolved?

[narrow garnet]

tysm

.close

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[uneven otter]

now idk how to find for what values

it would be inconsistent

Ik this wont have any solutions because second last row concludes that 0 = 1

[kind hawk]

whenever you divide by something, make sure that the something isn't zero

[uneven otter]

oh so since 0 = a - 2b

I cant divide row 3 by it?

[kind hawk]

if a-2b=0, then you can't divide by it

so there are two cases

either it is zero or it isn't

[uneven otter]

shit im confused

so do I split it into two cases

before doing that

but ltte say

if a - 2b neq 0

then the rref would be the same as what I got right

and the rref I have is incosistent too right

[shrewd wasp]

I guess you are right, it is inconsistent

[uneven otter]

brah so its correct

but

the question

is askin

for what values of a, b

[shrewd wasp]

$\forall \alpha, \beta \in \mathbb{R}$

[warm shale]

adzetto

[uneven otter]

why

[shrewd wasp]

because this is present regardless of the values of α and β

[uneven otter]

mhm alr

@thin star whats funny doe?

😭

do u think i fuked up

[kind hawk]

I am a bit worried that the program gives that rref

[uneven otter]

i checked on another

website

and got same

[kind hawk]

it should absolutely depend on the values of alpha, beta

[uneven otter]

as mine

[kind hawk]

instead of alpha and beta plug in zero

what do you get then

(you should obviously get a consistent system because the last column is the zero column)

[uneven otter]

wait

so all 0s

on the right side?

[kind hawk]

but yes that's what you have to do

yes

[kind hawk]

there we go

[uneven otter]

🤔

i dont get it

[kind hawk]

online calc can't handle variables correctly

[uneven otter]

so for one case im alrdy done right

[kind hawk]

yes

[uneven otter]

alright imma do case 2 rq and show u brb

[uneven otter]

uhhh another question

I was thinking about dividing row 3 by a

but i will have to split it into cases again right so is there a better way

[kind hawk]

just do it

you have two variables so it's not a huge surprise that you have to split it into cases two times

[uneven otter]

again doe

so a = 0 and a neq 0

[kind hawk]

yes

[uneven otter]

shit alr brb

[uneven otter]

im stuck

what should I do next]

wait could I do

row 1 - row 4

[kind hawk]

well now the third row is 0 0 0 0

[uneven otter]

and then say its inconsistent

[kind hawk]

and if a=0 and a-2b neq 0, then b has to be neq 0

[uneven otter]

no no

this is case 2 for a - 2b = 0

[kind hawk]

ah

[uneven otter]

so a = 0 and a - 2b = 0

[kind hawk]

well then b has to be zero

[uneven otter]

ohhhhh wb

[kind hawk]

so fourth row is also 0 0 0 0

[uneven otter]

row 4 + row 3

to get row 4 as 0 0 0 0 a - 2b

and row 3 as 0 0 0 0 a

[kind hawk]

you are overcomplicating this

[uneven otter]

ah mb

[kind hawk]

you can just plug a=0 in

and b=0

you don't have to do any more row operations

[uneven otter]

but

im confused

how did u conclude

that b = 0

from thsi

oh because

a - 2b = 0

[kind hawk]

a=0 and a-2b=0

[uneven otter]

and a = 0

ahhhhh

so

1 0 1 | 0

which is inconsustent

right

[kind hawk]

what

[uneven otter]

brah

like

1 0 1 | 0 is never true

u said

[kind hawk]

anything with a zero on the right is always consistent

[uneven otter]

to plug in b = 0

[uneven otter]

wait so ur sayinf

this matrix is consistent?

[kind hawk]

if a=0 and b=0, yes

but that is in fact the only case

[uneven otter]

so the answer would be for all a, b in R s.t a and b neq 0?

[kind hawk]

yes

wait

the and is wrong

a neq 0 or b neq 0

[uneven otter]

but

our final case

was

a = 0 and b = 0

because look at this

[kind hawk]

yes and the negation of that is a neq 0 or b neq 0

[uneven otter]

oh for inconsistency

for a b in R s.t a neq 0 or b neq 0

[kind hawk]

yes

[uneven otter]

for all a, b in R s.t a neq 0 or b neq 0?

[kind hawk]

again, yes

[uneven otter]

ty

.close

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[verbal sluice]

can anyone help me with how to prove a Cobb-Douglas function is homothetic

[royal basin]

what's a homothetic function

[fossil crag]

a homothetic function can be written as $f\circ g$ where f is monotonic and g is homogeneous (ie $g(sx_1,...,sx_n) = s^kg(x_1,...,x_n)$)

[verbal sluice]

yes, but now it's like I know u = $f\circ g$ but I dont know f and g , and I need to prove u is homothetic

[warm shale]

CCCCClaire

[fossil crag]

yes, well you have to find f and g such that it works

[verbal sluice]

here i should be 1

so can I say if (the partial derivative of tx1 / the the partial derivative of tx2 )= (the partial derivative of x1 / the the partial derivative of x2), then u is homothetic?

t is a constant

[warm shale]

rafilou2003

[fossil crag]

uh lemme check

you should be able to write the utility function as $(x_1^{\frac{\alpha}{\alpha+\beta}}x_2^{\frac{\beta}{\alpha+\beta}})^{\alpha+\beta}$

[warm shale]

rafilou2003

[verbal sluice]

ahhhh

and here the f is the inside function and g is ()^a+b

[fossil crag]

uh the opposite

f(g(x1,x2))

[verbal sluice]

ooh right

[solar trellis]

what's k

[fossil crag]

some integer

[verbal sluice]

and btw if I want to show u() is qusai-concave can I use its borded Hessian determinant ?

[solar trellis]

wth i've never heard of these words

[verbal sluice]

or just use its second order derivative and say if it is concave, then it is quasi-concave

[fossil crag]

yeah concave => quasi-concave

I think that's the case here

[verbal sluice]

thank u so much!

[obtuse pebble]

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[hushed jasper]

oops never mind

.close

[obtuse pebble]

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[rocky pebble]

Need help with this functional equation: f(x^2 + f(y)) = x•f(x) + y

[timid silo]

Don't use multiple channels

[obtuse pebble]

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[storm kayak]

Where do i even start?

[kind hawk]

do you know some other sets that generate A_n?

[storm kayak]

This is all information that was given

[kind hawk]

this can't be the only thing you know about A_n

[storm kayak]

ill give the full question for clarity:

[warm canopy]

What is your definition of A_n?

[storm kayak]

It is not defined anywhere so i dont really know

[polar fossil]

surely your book defines the symbol A_n at some point

[storm kayak]

Im trying to look for it right now

okay i found it A_n is the set of all even permutations

[warm canopy]

And what is an even permutation

[storm kayak]

definiton from the book says: A permutation that can be written as an even number of transpositions

[warm canopy]

So do you see how the s_i s_j generate A_n

[storm kayak]

not sure

[obtuse pebble]

@storm kayak Has your question been resolved?

[obtuse pebble]

@storm kayak Has your question been resolved?

[obtuse pebble]

@storm kayak Has your question been resolved?

[storm kayak]

<@&286206848099549185>

[obtuse pebble]

@storm kayak Has your question been resolved?

[wide star]

Why did the bot just stop asking?

[tepid mountain]

@wide star
It only asks for few times then it just stops

[slate zephyr]

I think it keeps asking but the waiting interval increases exponentially, right?

[timid silo]

im in hs ineed help with my algebra

[viral blade]

using this, what set would generate A_n for sure

[obtuse pebble]

@storm kayak Has your question been resolved?

[urban dawn]

i have to find AC

<@&286206848099549185>

<@&286206848099549185>

[spring steeple]

!occupied

[obtuse pebble]

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).

[slim cove]

Idk if you're still here, but #groups-rings-fields is probably a better place to ask future abstract algebra questions, more knowledgeable people will be there

Which part are you confused about though? Can you show that that set generates S_n?

[obtuse pebble]

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[wanton carbon]

How would I prove that the directrix of the quadratic function of the form y = ax^2 is vertical?

[obtuse pebble]

@wanton carbon Has your question been resolved?

[wanton carbon]

<@&286206848099549185>

also i want to show uniqueness

prove that there is no other directrix for this quadratic function

[shrewd wasp]

Let $P(x, ax^2)$ be an arbitrary point on the parabola $y = ax^2$. Let $F = (f_x, f_y)$ be the focus of the parabola $y = ax^2$, and let $d$ be the directrix by $y = c_1 x +c_2$. Use definition $$\mathcal{P} = { P \in \mathbb{R}^2 \mid d(P, F) = d(P, d) }.$$ Then find $f_x, f_y, c_1, c_2$ using the polynomial equation. This can be algebraically tiring. You can then use a reductio ad absurdum argument to prove uniqueness. Having found $F\left(0, \frac{1}{4a}\right)$ and $\left(c_1, c_2\right)=\left(0, \frac{-1}{4a}\right)$, using the same definition, assume $y=c$ and $c\neq \frac{-1}{4a}$ and show that there cannot be such a $c$.

[warm shale]

adzetto

[obtuse pebble]

@wanton carbon Has your question been resolved?

[wanton carbon]

One second though this is assuming we just have the equation of y = ax^2

this is part of a large proof

so we don’t know it’s a parabola yet

[wanton carbon]

basically solved but one second

[shrewd wasp]

What

If you want a complete proof, do you want to define/demonstrate algebraic elements, real numbers and many more?

I think it would be quite redundant

[wanton carbon]

ok fair enough

is there a way other than just the definition of a parabola to show that it can only have one focus

@shrewd wasp

[shrewd wasp]

Let $F_1 = (f_{1x}, f_{1y})$ $F_2 = (f_{2x}, f_{2y})$. Then again use the definition $d(P, F_i) = d(P, d_i)$. you'll have seen it $F_1 \equiv F_2$

[warm shale]

adzetto

[wanton carbon]

can u put this more plainly pls?

I’m an amateur

[shrewd wasp]

It should already have a single focus by definition, but if you want, assume that it has two foc such as F1 and F2. Write the equation for both focus using the definition. You will see that F1 and F2 must be the same point.

[shrewd wasp]

No problem, I'm studying engineering anyway.

[obtuse pebble]

@wanton carbon Has your question been resolved?

[obtuse pebble]

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[wet badge]

help is (4,38 )correct

[latent walrus]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[wet badge]

i jus need to verify my aswnser

(4)

^i didnt do the equation yet, since i want to check if i got the coords correct

[latent walrus]

aha, lets see
y=8x+4, m=-1/8
f'=-1/x^(3/2)=-1/8
x^3/2=8, x=4

[wet badge]

yes, after i get the x, do i plug it into the orginal equation

or the derived one

[latent walrus]

original if you want the y coordinate

[wet badge]

because when i plug into the orginal i get 32

oh okay tytyt

32 looks like a devious number, so i wasnt too sure

[latent walrus]

how do you get 32?

[wet badge]

y=8(4)+4

[latent walrus]

thats 36, but also thats not the equation of the graph my guy

[wet badge]

wait so what do i do

[latent walrus]

the graph has the equation f(x)=2/sqrt(x) +1

[wet badge]

oh my god i jus relised

thanks bro

[latent walrus]

np

[wet badge]

wait i dont wanna bother, but can you also explain this question

(you dont have to, but i was jus confued on it)

[latent walrus]

you can check the limits in either direction (+ or - infinity)
if f(x) is convergent to a value rather than infinity or - infinity then that can be considered an asymptote

[wet badge]

wait, but how would i use it with the limit method

lim h->0,( f(x+h)-f(x))/h

[latent walrus]

im not sure how youd use that formula specifically for this honestly, ive only ever used it for differentiation

ill do some reading though and come back if i find anything

[wet badge]

its okay i can give u a differnt problem ! (i was jus confused on how to factor out the x^2 for this one )

[latent walrus]

if you multiply the numerator and denominator by 1/x though then take limits to +inf and -inf it should come out cleanly though

i dont think you can factor out x^2 really, however, if you divide the numerator and denominator by x^3/2 and then take the limit it should be clearer

[wet badge]

ah okay tyty

[obtuse pebble]

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[timid silo]

Hello I am doing Statics homework which requires some Calc 3 vector math. I have unlimited attempts and I have been using them to check my answers. I am stuck on questions 4-9. I have my work but will include a picture of the question first.

** Note: It's just one question, but divided to 10 steps
Question:
Force F1 has a magnitude of 473N and force F2 has a magnitude of 514 N. h = 3.9 m,

1- Determine Force F1's x-component in N ? = 0
2- Determine Force F1's y-component in N ? = -216 N
3- Determine Force F1's z-component in N ? = -421 N
4- Determine Force F2's x-component in N ? = 210.7 N (incorrect)
5- Determine Force F2's y-component in N ? = 422 N (incorrect)
6- Determine Force F2's z-component in N ? =-206 N (incorrect)
7- Calculate the Resultant force's magnitude in N ? 693N (incorrect)
8- Calculate the resultant force's coordinate direction angle θx in Deg ? = 72 degrees (incorrect)
9- Determine the Resultant force's coordinate direction angle θy in Deg ? = 67 degrees (incorrect)
10- Determine the resultant force's coordinate direction angle θz in Deg ? = 155 degrees

so what i have done:
find the unit vector of F2 (direction vector of length 1)
multiply each component of the unit vector by the length of F2
obtaining F2's components using the unit vector

[timid silo]

u2 is a unit vector, of length 1, in the direction of F2.
when i multiply the magnitude of F with u2 i get the components of F2.

[trail cloak]

Alright are you given the value of h?

h = 3.9 m okay

[timid silo]

look, if i take the magnitude of F2's components i get 514 N,
and when i take the magnitude of u2 i get 1.

i think my teacher might be wrong so i want to double check

[trail cloak]

Hmm

For F2 I got this

$\bar{F_2} = \left \langle250.832, 376.25, -244.664 \right \rangle \mathrm{N}$

[timid silo]

okay, where do you think I might have gone wrong?

[trail cloak]

In the unit vector of F2

You used 8 instead of 8-2

[timid silo]

okay, accounting for z being -z direction, i used the components you found and it was correct

[warm shale]

VulcanOne

[trail cloak]

Alright

[timid silo]

i see what you are talking about, except my teachers notes confused me because of a problem similar to this where he did not do 8-2 = 6

[trail cloak]

I think from there you can redo your work and you'll hopefully get the correct answers

[trail cloak]

Yeah it can get confusing just writing one number

So I always write my ending coordinates and starting coordinates to tell myself where the vector was going and where it started

[timid silo]

thats how i started but i was following this example and he was trying to show us some short cut but I dont think i can use it for this homework problem. look at this example we did in class and how easy it was to find the position vectors (r)

thank you very much btw VulcanOne

[trail cloak]

Just writing the results is pretty confusing to me. Not sure where the shortcut is.

[trail cloak]

Also the notation is a bit off

For the magnitude, you'd use this

[timid silo]

true

[trail cloak]

$\norm{\overrightarrow{r_{AB}}}$

[warm shale]

VulcanOne

[timid silo]

yeah, the notation has been really messing with me. Since in Calc 3 we use one notation, and in Physics they used another one, and my engineering professor is just barely using any proper notation 😅 🥲

[trail cloak]

Wait are you structural engineering?

[timid silo]

Electrical & Computer Eng.

[trail cloak]

Ooooo

[timid silo]

got me like this🧟

[trail cloak]

Same lmao

[timid silo]

its a very humbling experience

[trail cloak]

Hope we make it into the big leagues soon

[timid silo]

are u doing structural?

[trail cloak]

Yep. Civil engineering. But structural is like 80% of the entire career so yeah :)

[timid silo]

That sounds awesome and also very humbling 😅

[trail cloak]

It is indeed

[timid silo]

so i was wondering what points u used for the starting coordinates

[trail cloak]

I used the point where the forces where exiting

Point A

And the other 2 points I used as the end points

Since direction matters

And the forces have their arrows pointing that way so ye

[timid silo]

if i use A (0,2,3.9) how is it when i get the vector AB, it is <0,-2,-3.9>? when the z point of A was positive?
i am just confused on where i need to do the book keeping

ooooh wait i see it now....i was counting point A as (0,0,0) idk why

i was trying to use free body diagram technique

[trail cloak]

Well it can get messy if you relocate your origin

[timid silo]

so i just keep the origin at the original given one? no shortcuts?

[trail cloak]

Yep

It's better that way

[timid silo]

okay

[trail cloak]

You won't lose track easily

And you can look at the diagram and it will be your reference

[timid silo]

You're right and I think everything messed up when i found the magnitude of AC = <4, 8,.-3.9>, since it should be AC = <4, 6, -3.9>

that little difference ruined everything after that

i swear these little things are going to keep me from getting A's this semester

[trail cloak]

You'll get an A I'm sure

Keep things as simple as possible

You'll get the A+ in the bag

Also time yourself and know where you are messing up and try to fix them

No shortcuts until you understand the process

After that you will shortcut automatically

[timid silo]

thats some great advice that didn't know i needed...

I just finished the homework and got everything correct, thanks again bro

idk if ur accepting FR's but u should add me if you ever need another eye or helping hand!

.close

[obtuse pebble]

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[untold cove]

Is this correct?

[cunning breach]

You should write negative sign in step 2 instead of absolute value. $$\int_{-1}^1 -\frac{3}{4}x+\frac{3}{4} dx - \int_{1}^2 -\frac{3}{4}x+\frac{3}{4} dx$$

Since $$\int_{1}^2 -\frac{3}{4}x+\frac{3}{4}x dx$$ has a negative area.

[warm shale]

Good

[cunning breach]

Hmm, the value I got is 9/8

[untold cove]

Oh okay

Thanks

Tbh k don’t understand half of that stuff

[cunning breach]

From step 4 to step 5, you forgot your negative sign.

[untold cove]

My teacher just suddenly started writing wrong equations and I got confused lol

Thanks!

[cunning breach]

Welcome.

[untold cove]

Why is there 3/4x dx? Thought it was 3/4 dx…?

@cunning breach

[warm shale]

Good

[cunning breach]

My bad, since I copy-pasted it.

[untold cove]

Oh okay, thought I was trippin

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[obtuse pebble]

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[proper kelp]

if f(x)=1+xlnx for (0,5], what is its minimum and maximum value within the domain

[hidden compass]

Min/max should ring you a bell about derivatives, right?

[proper kelp]

right, so the minimum i found to be 1/e by simply equation the derivative to 0

but for the maximum, im not tpoo sure

intuitively, i would say x=5 but without proper working out

[median spoke]

f'(x) = lnx + 1

f'(x) = 0, x = 1/e as you say

so you should test x = 0, 1/e, 5

[proper kelp]

ah

so how would i word this

as x increased, f'(x) increase, therefore the greatest x would yield the greatest f(x)

but this is kinda weird to write in a maths problem

[median spoke]

it wants you to explain?

[proper kelp]

no

but i think i should have some working explaining why its 5

[median spoke]

You always test the highest and lowest values in the domain

so 0, 5, and then the value in the domain where f'(x) = 0

[proper kelp]

well i cant test 0 since lnx

[median spoke]

0 technically isn't in the domain, but as you approach 0 from the right you get 1

[proper kelp]

the limit approaches 1?

[median spoke]

it does, but I guess that's not obvious without looking at the graph

but the derivative is negative between 0 and 1/e

so the minimum is def 1/e

so yea it's 5 because that's the one with the greatest value of the 3 significant values

[proper kelp]

i see

can i ask another question?

[median spoke]

yes

[proper kelp]

x^2+xy+y^2=1

what is the max and min y values possible?

i found the derivative to be -(2x+y)/(2y+x)=0

[fervent cradle]

i think you can just rearrange for y explicitly

like, you can get y = f(x)

and then go normally

[proper kelp]

how would you do that

[fervent cradle]

ok suppose you had 1 + y + y^2 = 1, how would you solve for y

[proper kelp]

quadratic

[fervent cradle]

right

[proper kelp]

or just factorise

[fervent cradle]

so just do that for x^2 + xy + y^2 = 1

i mean ok

you don't need the quadratic formula for this, it's actually nicer to use completing the square when you have algebra i find

maybe slightly simpler

but both work

[median spoke]

(1 - 2x - y)/(2y + x)

getting that for derivative

wait no

nope you're right

[fervent cradle]

ok wait i'm dumb

[fervent cradle]

given this

for now let's just multiply through by the denominator

[proper kelp]

wait i meant y'=that

[fervent cradle]

and then you just have -(2x+y) = 0

oh

[proper kelp]

yep

yeah its still the same thing

[fervent cradle]

i mean ok but you want y' = 0

yeah

so yeah you have -(2x+y) = 0

and then you can just substitute that back into the original thing

and then it's just quadratic again

that's probably how you're intended to do it

[proper kelp]

i tried that and got y= plus minus 2/root7

which was not right on the graph

[fervent cradle]

oh

ah i see your problem maybe i think

so let's just do it

y = -2x and x^2 + xy + y^2 = 1

so x^2 - 2x^2 + 4x^2 = 1

so 3x^2 = 1

so x = +-1/sqrt(3) and then y = +-2/sqrt(3)

simple sign error

you probably had like y = 2x instead of y = -2x

[proper kelp]

oh yeah you're right

[median spoke]

getting 3x^2 = 1

[fervent cradle]

yup

[median spoke]

x = ±1/sqrt(3)

[proper kelp]

ok so im having trouble on which to substitute, x=-1/2 y or y=-2x

[fervent cradle]

either works

[median spoke]

shouldn't matter should it?

[fervent cradle]

solve for either x or y and then you can just use y = -2x to get the other one

[proper kelp]

but would x= something y be faster

[fervent cradle]

sure whatever

you'd have to deal with fractions tho

so i didn't

[proper kelp]

i see

so i simply sub x=plus minus 1/root3 into the original equation

and find y

[fervent cradle]

no

[median spoke]

yes

[fervent cradle]

i mean ok you can

but it's much simpler to use y = -2x

once you have that

[proper kelp]

oh true

also since x^2+y^2+xy means the values of x and y can be interchanged, would y just be the same answer

[fervent cradle]

yyyes

like ok

[proper kelp]

right what is this called again?

[fervent cradle]

y is maximum at x = -1/sqrt(3), and then y has max value 2/sqrt(3)

[proper kelp]

actually no scrap what i said

[fervent cradle]

and then x is maximum at y = -1/sqrt(3) and x has max value 2/sqrt(3) there

[proper kelp]

oh wait nvm

[fervent cradle]

so like, you can swap them in that respect

[proper kelp]

yeah thats what i meant, im confusing myself

actually

wait why is y max at -1/sqrt3 instead of the postive version

[median spoke]

graph looks like this for reference

[fervent cradle]

ok

so we have y' = 0 when x is either 1/sqrt(3) or -1/sqrt(3)

for x = 1/sqrt(3), y = -2/sqrt(3)
for x = -1/sqrt(3), y = 2/sqrt(3)

so one of these is a minimum and one is a maximum

[proper kelp]

oh i see

ty i understand now

🙂

.close

[obtuse pebble]

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[obtuse pebble]

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[proven zephyr]

Suppose there are 2 real integers with a difference of 10. The smaller integer is a multiple of 3 whereas the other is a multiple of 7. If the sum of all the prime factors of each of the integers is 17, find the value of those 2 integers.

[proven zephyr]

So.... the sum of the other prime factors would be 17 - 7 - 3 = 7

Which limits the possibilities

But I don't know how to start

[tender stratus]

doesnt this imply both must be having 5 as a factor

because otherwise you dont get 10 as the sum

3 and 7 already taken, and 2 + 8 isn't possible

[proven zephyr]

But you can use 3 for the first integer and 7 for the second integer

it can also be
2, 3, 5

[tender stratus]

hmm I guess that's possible

wait

the sum of the prime factors is including both right?

[proven zephyr]

yes

prime factors of the larger integer + the smaller

<@&286206848099549185>

[obtuse pebble]

@proven zephyr Has your question been resolved?

[errant lark]

Do we consider the distinct prime factors or count them as many times as they appear in factorisation of a number?

[proven zephyr]

distinct prime factors

like for example

2^5 * 3^7 would be 2 + 3

= 5

[errant lark]

Okay.

[errant lark]

But if other number had 2, would that count as separate?

[proven zephyr]

Brb 5 min, my food came

no

wait yes

i mean yes

[errant lark]

Okay.

[obtuse pebble]

@proven zephyr Has your question been resolved?

[errant lark]

Are you certain that there is an answer? I think i ruled out almost all of them. Of course, i could have made a mistake.
Anyway, confirm if there is indeed a solution.

[proven zephyr]

Yes there is i found one on the internet

But i can’t understand it

wait

i will translate

[errant lark]

Sum of other prime factors would be 7. Right?
I can't see any possibility other than (2+5) or (2+2+3).
Both of which are not possible.

[proven zephyr]

oh wait i did a typo

[errant lark]

😐

[proven zephyr]

but yes

i wrote it as 17 - 7 - 3 = 10

edited tho

_ _

For example:
the two numbers are (x, y)
x = 7m
y = 3n

since the difference is 10 and x>y
then:
7m - 3n = 10

the values of m and n can be determined by reversing Euclid's algorithm
7 = 2×3 + 1
1 = 7 - 2×3
10 = 70 - 60

thus obtained
m = 10
n = 20

Thus, the general solution
m = 10 - 3t
n = 20 - 7t
x = 70 - 21t
y = 60 - 21t

The satisfying pairs of two-digit numbers are ={(28,18),(49,39),(70,60),(91,81)}

Note that the sum of all prime factors of x and y is 17, hence 17=3+p+q+7.
Then p+q =7, so the only primes p,q that satisfy are 2 and 5.

So, it is clear that among the pairs x,y, the only ones with prime factor 5 are x=70 and y=60. So, the sum of these two numbers is
x+y = 70+60=130.

i don't understand how they got m = 10 and 20 = n

[errant lark]

70 = 2* 5 * 7
60 = 2 * 2 * 3 * 5

Sum = 2 + 3 + 5 + 2+ 5 + 7 = 24

Not 17

[proven zephyr]

wait yeah

what

[errant lark]

Or you told me wrong.

[errant lark]

Here.

[proven zephyr]

Okay I'm not sure now.

[errant lark]

According to what you said, there is no solution.

[proven zephyr]

Hm.... okay, maybe it doesn't count as separate.

[errant lark]

Other than that, approach is really simple.

You see that GCD of 3 and 7 is 1.

That means there are infinitely many numbers of the form 3n and 7m such that 7m - 3n = 10

Now, you just solve that using euclidean or diophantine equation.

You get general solution as m = 10+3k, n = 20 + 7k, where k is any integer.

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[proven zephyr]

.reopen

[obtuse pebble]

✅

[proven zephyr]

forgot to react mb

[errant lark]

Do you want me to show what i mean by euclidean method?

It's just reverse GCD process.

[proven zephyr]

Sure, if you're okay with it.

I'm not sure i've learnt it.

[errant lark]

I'm pretty sure you have. At least the forward part.

So, we calculate GCD of 3 and 7.

$$ 7 = 3\times 2 +1 $$
$$ 3 = 1 \times 3 + 0$$

[warm shale]

Enemagneto

[errant lark]

GCD = 1

Clear as much?

[proven zephyr]

I can see that GCD = 1

[proven zephyr]

But what is this for?

[errant lark]

You have never calculated GCD using this method?

How do you calculate GCD?

[proven zephyr]

I-

last time i did that it was at 4th grade

ima just go and relearn this after this

[errant lark]

Sure.

Also, learn how to represent GCD of a and b in the form ax + by, where x and y are integers.

That's all it is about.

[proven zephyr]

Okay!

Btw, an you continue with the explanation? I'll go back understand it after I've learned about it.

[errant lark]

Okay. So, basically you try to find such two numbers m and n such that 7m - 3n = 1.

We know that there are such numbers because 1 is GCD of 7 and 3.

It's easy to see in this case that m = 1 and n= 2 works for us.

So, we have 7(1) - 3(2) = 1

but we need 7m - 3n = 10. Right?

[proven zephyr]

yes

[errant lark]

So, we multiply this equation with 10.

We have 7( 1 * 10) - 3 (2 * 10) = 10

Which is just 7(10) - 3(20) = 10

Lucky for you, your answer is already there. 70 and 60. But i'll explain further for general solution.

So, once you have found one pair of m and n that satisfies
7m - 3n = 10, you can find all the infinite solutions of this equation where m and n are integers.

This is something you can also think on your own. Would you like to try or should i move on?

[proven zephyr]

You can move on.

[errant lark]

Alright. So, key to finding all the solution is to imagine that we'll have different m and n but the RHS will always be 10.

So, basically you can say that both the terms 7m and 3n are increasing/decreasing by equal amounts so that their difference stays equal i.e. equal to 10.

So, we have something like this:
(7m + l) - (3n + l) = 10

Our m and n are currently 10 and 20 respectively. Remember that.

[proven zephyr]

okay

[errant lark]

You can take any value of l here and equation will satisfy.

But if we take such a l which is divisible by 7 then we can factor 7 out from l.
So, we'll have something like:
7(m+ k) - (3n + l) = 10

So, you see - Here, we have kind of found another integer m' = m+k which will satisfy our equation. Only issue is that it's not our solution yet as we have not found a corresponding n.

So, think about how we can get n'. Well, for that l has to be divisible by 3. Then we can factor 3 out from it.

So overall, if l has 3 and 7 both as its factor, we will get new m' and n'.

Makes sense ?

[proven zephyr]

Why does n need to have 3 and 7 both as its factor?

[errant lark]

Not n

l

[proven zephyr]

ohh

okay i get it then.

[errant lark]

I fixed the typo.

Great.

So, l will basically have to have 21 as its factor i.e. l will be of the form 21k.

Now, we are almost there.

Let's put it in the equation.

(7(10) + 21k) - (3(20) + 21k) = 10

(7(10) + 7(3k)) - (3(20) + 3(7k)) = 10

7(10+3k) - 3(20+7k) = 10

Where k can be any integer.

So, m = 10+3k & n = 20+7k

Just for fun, let's try putting some value of k.

Any integer.

[proven zephyr]

3

[errant lark]

Okay

m = 19 & n = 41

7(19) - 3(41) = 133 - 123 = 10

So, you have found all infinite solutions.

Now, work on m and n as you need them to have your required factors.

[proven zephyr]

Do I just test them one by one or is there a faster way? (Sorry.... I'm really bad with number theory)

[errant lark]

Well, fun thing about number theory is that it's very intuitive and realistic. Who doesn't feel cozy with the simple natural numbers so hang in there.

[errant lark]

I'll give you a few pointers.

Look at your numbers.
You have 7(10 + 3k) and 3(20+7k).

You already have 3 and 7. So, remaining sum is 7.

Two possibilities:
2+5 OR 2+2+3

[proven zephyr]

But we don't count them as separate.

2 + 2 + 3 would be 5 no?

Since we're counting 2 once

[errant lark]

Ah. Yes. Still in mindset of original wording of the question. Lol

Only one possibility. 2+5

That makes it way more simple.

So, either 10+3k would be divisible by 2, and (20+7k) would be divisible by 5, or other way around.

Now try to think of k values which will make that possible.

Should be easy?

[proven zephyr]

0, 10, 20, .... , 10 * n

and so on

[errant lark]

Yes

All of those will give correct answers.

[proven zephyr]

So then I just test them one by one?

wait... all of those give correct answers?

so-

huh?

ima just take a random k like.. 40

7(10 + 3k) - 3(20+7k) = 7(10 + 120) - 3(20 + 280)
= 7(130) - 3(300)
= 910 - 900

Prime factors of 900: 2, 3, 5

Prime factors of 910 = 2 x 5 x 7 x 13 nvm

so just.... 0

okay

thanks

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[proven zephyr]

t!rep @errant lark

[sinful light]

marvei has given @errant lark a reputation point!

[errant lark]

Yes

Lol

[obtuse pebble]

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[timid silo]

can someone help me understand this

[timid silo]

discrete math

[onyx gust]

I am struggling to understand the solution for case 2, mainly at why sqrt An + sqrt L disppear when they are showing the proof.

[timid silo]

kittie, i have this help-10 covered

[golden sand]

! occupied

[obtuse pebble]

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).

[timid silo]

post on help-forum

[blazing forge]

how formal do you have to be?

[timid silo]

wym how formal

[blazing forge]

i'm not sure i could type out a perfect natural deduction proof

but if you just need to reason you way to ~q with the premises

[timid silo]

as long as i understnad we good

[blazing forge]

i can help with that

pog

[timid silo]

yes yes

[timid silo]

no u

[blazing forge]

okay. so you want to somehow conclude ~q. the only premise with ~q in is is s v ~q. in order to get ~q from this, we need for ~s to be true

basically you can just reason backward, if you want

[timid silo]

wait why do we need ~s for ~q to be true

[blazing forge]

you know that s v ~q is true. if you want this premise to force ~q to be true, you have to know that ~s is true. i guess you can also see this by noting ~a v b is equivalent to a --> b. so s v ~q is equivalent to ~(~s) v ~q, which is equivalent to ~s --> ~q

so if we know ~s is true, then we get ~q is true

since the only way for ~s and s or ~q to both be true is if ~q is true

[timid silo]

damm im lost asf

[blazing forge]

what are the argument forms in Table 2.3.1

this will likely help

[timid silo]

im looking at elimination

p V q

~q

[blazing forge]

pog yeah, you can use elimination

[timid silo]

therefore p

but lets swap this

[blazing forge]

you know s v ~q is true, so if you know ~s is true, you can conclude ~q is true

via elimination

[timid silo]

s V ~q
~s
shouldnt it be q instead of not q?

[blazing forge]

this is just saying for arbitrary premises p and q. ~q is an arbitrary premise

it would be better if they said a v b, ~a, therefore b

then you could say you have s in place of a and ~q in place of b

[timid silo]

okay so in any case, if the premise has not in elim

then the conclusion is also not

got it

i dont need tto understand why fuck it, i got a test in 30 min

[blazing forge]

you could have something like (p --> r) v (q --> s --> t) and if you knew ~(p --> r), then you could conclude q --> s --> t

with elimination

[timid silo]

right

i fuckin wish this prof would give hw

[blazing forge]

pog. so, if we do the reasoning backward method, we want to know ~s is true in order to do elimination on b. to conclude f.

so we have to somehow get ~s from the premises

e. tells us that if we know ~p and r, we will then get ~s (by using modus ponens)

so now we just need to know ~p and r is true from the premises

does this sort of make sense so far

[timid silo]

yes

[blazing forge]

pog. now a. tells us that ~p v q --> r, so if ~p is true, then ~p v q will be true (by generalization), so then modus ponens will tell us that r is true.

so in fact, we just need to know that ~p is true in order to know r is true (and therefore ~p and r is true, by conjuction)

c. tells us that ~t is true, and modus tollens on d. will tell us that ~p is true

[timid silo]

yes

[blazing forge]

then you just have to write it all in the other direction...

[timid silo]

help me understand this

how does ~t being true

tell us ~p is true?

[blazing forge]

d. says that p --> t, and you know that ~t is true, so then modus tollens says ~p is true. or do you mean why does modus tollens work?

an example might help. it is true that if it is rainy, then it is cloudy. so rainy --> cloudy. if somebody tells you that its not cloudy, then you can use this fact to conclude that it is not rainy

this would be an example of applying modus tollens in a case where its more intuitively clear

[timid silo]

yes got it

[blazing forge]

because if it were rainy, then it would be cloudy. but we are assuming its not cloudy

[timid silo]

are u able to talk rn?

[blazing forge]

no sorry

[timid silo]

😦

[blazing forge]

next to my gf who is in a call

[timid silo]

ill send u 5$ enough to buy her chocolate 😉

" yo babe if i talk with this guy, you get chocolate, deal?"

[golden sand]

@timid silo you are breaking the rules here

[blazing forge]

so ~t will modus tollens d. into ~p. then this will generalize to ~p v q, which modus ponens on a. into r, so then ~p and r are true. conjuction gives ~p and r is true. modus ponens on e. tells us that ~s is true, and then elimination on b tells us ~q is true

would be writing our backward reasoning forward

[timid silo]

am i?

sorry!

mod didnt say anything bout it yesterday

[timid silo]

u got time for 1 more?

[golden sand]

#rules message

[blazing forge]

no because i assume its either going to be slowly telling you the answer again or we will run out of time before your test

use the process i've conveyed to you and your inbuilt human rationality to try to reason through the next one

[timid silo]

"Do not offer money for doing homework assignments, and vice versa" this is not for doing hw, this is for teaching me

idk if its still breaking the rule

[blazing forge]

i feel like you just need much more practice doing these things to do well for your test

i can't help you enough in 20 minutes or whatever

[timid silo]

yes i was studying yesterday

and tutor.com kept putting me with no audio

[blazing forge]

did you do any on your own :o

[timid silo]

nope cus i had more things to go over bahah

okay how bout this

ill do the next one

and u check it

[blazing forge]

sure

hopefully you can do at least one on your own before the test

i believe in you

[timid silo]

okie kinda stuck now

[blazing forge]

nice. so you have p and you want to get p and s in order to get t using modus ponens on c.

so you just need to somehow get s

there is only one premise that allows you to conclude s, which one is it?

[timid silo]

~q --> u ^ s

[blazing forge]

pog

so you just need to know ~q is true and apply modus ponens

[timid silo]

~q --> u ^ s
~q
therefore u ^ s

modus ponens

[blazing forge]

pog. now you can specialize to go from u ^ s to s

[timid silo]

yeah thats the part im stuck now

wait

so s

p ^ s --> t

idk

[blazing forge]

yea, you have p from the elimination you did earlier, so you can do conjunction to go from p and s to p ^ s

then modus ponens to get t

[timid silo]

ohh i can do that?

[blazing forge]

yeah

its called conjunction here

if you have p,q then you can get p ^ q

[timid silo]

what happens to

u ^ s?

[blazing forge]

theres one on the table called specialization

if you have p ^ q, you can get q (or p)

you have u ^ s, so you can get s

[timid silo]

right

damm

okay to p ^ s --> t

[blazing forge]

yeah. then you're done

[timid silo]

whaaa

fr?

shii so if i remember the table im gucci

i know these things intuitively but i might need to write down after i use it

on the test

[blazing forge]

you got p from elimination on a, and then you got s from specialization on u ^ s. so you get p ^ s by conjunction, then modus ponens p ^ s to get t

yeah

gotta know all the rules and their names i assume

part of being in a logic class

[timid silo]

can i add you?

i gtg

[blazing forge]

idk probably not i don't wanna do tons of logic problems

[timid silo]

bahahah

[blazing forge]

i already did that 1-2 years ago

[timid silo]

😦

me sad

[blazing forge]

[timid silo]

Appreciate the help bro

[timid silo]

Imma ace it

[blazing forge]

you can do it

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[radiant moss]

I have to turn this expression in the from asin(bx+c), I know I have to use Addtion thereom , but I'm not sure how. I'm thinking of using sin(x-y) = sin(x)*cos(y)-sin(y)*cos(x) but I am not sure how

[tardy epoch]

can you take a picture/screenshot of what you learned as "addition theorem"

[modern cedar]

is this correct ? i didnt replace sin at the end but i get another answer on photomath and the math solvers
its integral of dx/rad(144+x²)

[obtuse pebble]

@radiant moss Has your question been resolved?

[tardy epoch]

apply the sum formula to cos(3x) and sin(3x)

use 3 = 2 + 1

[obtuse pebble]

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[strong wigeon]

yo

[strong wigeon]

How is this X less or equal to 3?

I know R\ makes (3) into [3], but does it also make the (->) into [<-]??????

-> means infinity

<-

[obtuse pebble]

@strong wigeon Has your question been resolved?

[keen garnet]

It’s say’s that x is real number inferior to 3 I think

[obtuse pebble]

@strong wigeon Has your question been resolved?

[strong wigeon]

No bro its equal or less

But ty for attempting

[blazing forge]

its saying $x \in \mathbb{R} \setminus (3, \infty)$

[warm shale]

Ves

[blazing forge]

if i understand correctly

that's the set $(-\infty, 3]$

[warm shale]

Ves

[blazing forge]

and $x \in (-\infty, 3]$ if and only if $x \leq 3$

[warm shale]

Ves

[blazing forge]

or at least, that's how i'm interpreting it

the notation is not familiar to me

[keen garnet]

For me when I say inferior it’s inferior or equal

[strong wigeon]

You Are right but im confused Why infinity went to negative

Also since that E symbol say X before

I think it means X must be 3 or less

[strong wigeon]

Ok

[blazing forge]

it means that x is in R, but x is not in (3, \infty). the set (3, \infty) is the set of all numbers y such that 3 < y

so its saying that x is a real number that does not satisfy 3 < x

so that means that x satsfies x <= 3

and then thats the definition of the set (-\infty, 3]

[strong wigeon]

Ok thanks

if there is R \ the infinity changed direction?

[blazing forge]

in this case, but i don't think that's how you should think of it

you should unravel the definition of the set to see what its saying

[strong wigeon]

How i think of infinity ?

[strong wigeon]

Ok

Ty

[blazing forge]

it seems like your first thought was "if you have R \ before an interval with an infinity in it, then this is equal to an interval with an infinity on the other side". and this is basically always true, but its probably better to understand why it happens than to try to memorize this

[strong wigeon]

👍

.close

[obtuse pebble]

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[verbal niche]

Oh

.reopen

.close

[trail cloak]

Next time don't delete the original message

Please open a new channel

[obtuse pebble]

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[glossy temple]

help

[obtuse pebble]

Please don't occupy multiple help channels.

[obtuse pebble]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem, and is confusing for helpers, making it more difficult for them to answer your question. We ask you to please be patient, even if your channel has not received much attention. If you are having a problem with a helper, please report it to <@&833154869451292722> rather than closing the channel and attempting to seek out another helper.

@glossy temple Has your question been resolved?

[tardy epoch]

Closed your other channel

[obtuse pebble]

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[daring cairn]

Hi, I Had this question from a previous year test as im studying and im not sure if i got the right answer as the one filled in is wrong but no right answer is written.

using the digits0-9, find the number of natural numbers that are even, have distinct digits and are less than 5000

I got 1373

[daring cairn]

2* 8 * 7 * 4 = 448
2* 8 * 7 * 5 = 560
8 * 8 * 5 = 320
8 * 5 = 40
5 = 5
total = 1373

[obtuse pebble]

@daring cairn Has your question been resolved?