help-43

[strange pendant]

dot product between two vectors in r3 gives a scalar

[muted elbow]

yes

[strange pendant]

how is that an orthogonal projection

if its a scalar

[muted elbow]

idk what is orthogonal projection

[strange pendant]

.

[muted elbow]

maybe projection is not the term used in the book you read

idk how to explain it

you gotta watch a video on the geometrical meaning of dot product

then you will get it

[strange pendant]

.close

[compact pewter]

Channel closed

Closed by @strange pendant

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[deft tangle]

L is finte so value of A will be?

I tried to use binomial expansion

( a - |a| \left(1 - \frac{x^2}{2a^2} - \frac{1}{8} \frac{x^4}{a^4} \right) )

[boreal girder]

Andy

[deft tangle]

What should I do next?

[latent ridge]

try rationalizing it

[deft tangle]

Which term?@latent ridge

[flint oar]

Rationalize it with a+√(a^2-b^2)

[compact pewter]

@deft tangle Has your question been resolved?

[stone wing]

Or expand the terms

[deft tangle]

@stone wing

[jagged blade]

Is L'Hopital not an option?

[stone wing]

L hospital 4 times?

[jagged blade]

I think 2 might be sufficient

Since the numerator will clean up

[stone wing]

Yea didn't see it

[jagged blade]

I'm also tempted to use a trig sub here

[stone wing]

x=asin§ ?

[jagged blade]

Don't just give answers without explanation

[stone wing]

Sorry I'm new

[jagged blade]

The idea is to help them solve it themself

[jagged blade]

Yeah

[stone wing]

So for the limit to exist the x^2 terms must vanish

[stone wing]

I am myself a beginner actually.

[compact pewter]

@deft tangle Has your question been resolved?

[deft tangle]

denominator term x^4

[deft tangle]

Nobody is looking at it and giving me further response what I do next?

[stone wing]

Yea x^2 terms must vanish

So what must be the value of a?

[stone wing]

.

[deft tangle]

.close

[compact pewter]

Channel closed

Closed by @deft tangle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[carmine garden]

I don'

[carmine garden]

I don't get how 3 implies 2

[kind viper]

they don't show 3 => 2 outright

this solution does it as 1 => 2 => 3 => 1

[carmine garden]

I see

okay that works then

thanks!

[carmine garden]

I don't think there's a simple way to show that, is there

2 to.3 that is

[kind viper]

?

did you mistype or what

[carmine garden]

yes

3 to 2

[kind viper]

3 => 2 is... eh.

like without going through 1 somehow?

yeah i don't see a way immediately

[dusky nymph]

you could argue somewhat similarly to how they did 3==>1, like as follows:

[carmine garden]

what I see is we can say let t be in their intersection, then for any vector v

and yeah, that should work

[dusky nymph]

suppose x is a common element of v + U and w + U, say
x = v + u1 = w + u2
then v + u1 is contained in w + U
if u is any element of U, then
v + u = v + u1 + (u - u1) is contained in w + U + (u - u1), and the latter is just w + U

[carmine garden]

got it

Thanks!

[dusky nymph]

er, i might have garbled that a bit, but that's the general gist

[carmine garden]

got it

thanks

[dusky nymph]

there fixed typos in the last line

so that shows v + U is contained in w + U
and the reverse containment is shown identically

[carmine garden]

got it

thanks

[dusky nymph]

yw

[carmine garden]

.close

[compact pewter]

Channel closed

Closed by @carmine garden

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[torpid carbon]

yo what’s the deal with double derivatives

[torpid carbon]

second

[unkempt sail]

Do you have a specific question in mind

[torpid carbon]

yes i’m getting this map of stationary points and setting f’(x) as 0

and then checking with second

yo find a maxima or minima

when do you use this

and how does this work

[unkempt sail]

Ah I see

Second derivative usually gives the shape of the curve i.e convexity/concavity of the curve

[torpid carbon]

okay… do in what scenarios would you need to utilise the second derivative

[compact pewter]

@torpid carbon Has your question been resolved?

Channel closed

Closed by @torpid carbon

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[silent wigeon]

Are my answers fine

[silent wigeon]

Allowed by mark scheme BUT mine are in minus

[icy nymph]

why would u do it that way 😭

[icy nymph]

only one of your solutions is right

it should be 3.73, not -3.73

[silent wigeon]

?? I used the quadratic equation

[stone wing]

Yea you could just take the root lol

[icy nymph]

^^

[icy nymph]

just square root both sides

[compact pewter]

@silent wigeon Has your question been resolved?

[storm lichen]

(x-2)^2 = 3
|x-2| = √3
x = √3 + 2
x = -√3 + 2

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[hushed magnet]

,rotate

[hushed magnet]

whats 6.20

ok

yes you can basically just copy 6.20

instead of the straight line you have your given cont function between two points

[compact pewter]

Channel closed

Closed by @copper loom

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[pallid gorge]

For part c the answer is no when i got hes

Yes

[kind elm]

Well the inflection point is technically the point at which the concavity of the function changes

[pallid gorge]

Ya

[kind elm]

you need to ensure that f'' is of different signs on either side of your inflection point

which it is not, it's always non-negative

for a simpler example, x^3 does have an inflection point at x = 0 but x^4 does not

despite both second derivatives being 0

(it can be graphically verified too)

[pallid gorge]

wait i am confused

so i can sub in

[kind elm]

.

[pallid gorge]

x = 0

and x = 2?

and then if theyre different signs

[kind elm]

not always but yeah

[pallid gorge]

then x = 1 is an inflection?

[kind elm]

it doesn't have to be in a space of 1

[pallid gorge]

oh

[kind elm]

it could be within any distance

[pallid gorge]

what woul dbe the easiest

[kind elm]

in your case you don't have to do anything

since f''(x) = 12(x - 1)^2 >= 0

[pallid gorge]

my teacher did

the third derivative

[kind elm]

yeah that's another check

third derivative should be non-zero

[compact pewter]

@pallid gorge Has your question been resolved?

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[forest token]

not able to start

[muted elbow]

in near proximity of q1 the effect of q2 is negligible

same for q2

using this you can find the sign of charge

there is an equilibrium point to the right of q2

that means q1 has higher magnitude

[forest token]

did not understand this

[forest token]

just because there is an equilibrium point why does that mean q1 has higher magnitude?

[muted elbow]

E field intensity due to a particular charge ∝ 1/r^2

[forest token]

yeah

[muted elbow]

when r→0 for q1 this is a very big number

for q2 its some small number

so the total field can be approximated as that of q1 only

since q2 has negligible contribution

[forest token]

isnt it going to infinity for q2 also?

[muted elbow]

near q1 what is the distance from q2

[forest token]

infinity?

[muted elbow]

no

its not clearly mentioned but

q1 and q2 are at a finite separation

[forest token]

okay..no distances are mentioned tho...

what do i answer for this

[muted elbow]

its some finite distance

we dont need it

lets call it L

[forest token]

ok

[muted elbow]

so when the distance from q1 → 0 the distance from q2 → L

[forest token]

yeah

[muted elbow]

the magnitude of E field of q1 tends to infinity

while for q2 its some finite quantity

so its negligible in comparison

[forest token]

ohk yeah

[muted elbow]

so towards right of q1 we have - ∞

[forest token]

so q1 will be bigger but how to comment on the signs

[muted elbow]

that means direction is towards left

then what is the charge of q1

[muted elbow]

no

this does not comment on magnitude of q1

this is for sign

[muted elbow]

sign of charge

[forest token]

but E inversely proportional to r^2 and directly proportional to q..q1 is still having an impact at a distance L from q2 dosent that mean q1 is bigger

[muted elbow]

yes but

q1 and q2 are not given to us

we just assume them to be some finite quantity

i am just changing the distance

consider everything else as constant

[forest token]

negative?

[muted elbow]

like lets say we are 10^-10 m to the right of q1

[muted elbow]

yes

[forest token]

okay

how do i comment on magnitude then

[muted elbow]

you dont have to

just comment on relative magnitude

for q1

we will have something like 1/(10^-10)^2

while for q2 we will have 1/(L-10^-10)^2

which is bigger

although yes you can set L such that its actually bigger for q2 but there will always exist a neighbourhood around q1 where effect of q2 is negligible

[forest token]

huh why are we doing inversely proportional to 1/r^2 when we are talking about magnitude of the charge and not electric field

[muted elbow]

we are doing sign rn

have you understood doing sign?

and why it works

i can help with magnitude as well

what is the sign of q2

[forest token]

oh ok

[forest token]

i think ive got it..electric field to right of q1 goes to -infinity when q2 effect is negligible so it will be negative?

[muted elbow]

yes

it is attracting towards its right

well left too

but you get what i mean

[forest token]

yeah

[muted elbow]

now what is the sign of q2

[forest token]

hmm left of q2 it goes to -infinity so it is trying to attract

so positive

[muted elbow]

repulse

but ye you are right

[forest token]

repulse meaning?

[muted elbow]

repulsion

[forest token]

why will it repel

[muted elbow]

the test charge is assumed to be positive

q2 is positive

so it will repel it

repulse is wrong i think

forget that

[forest token]

ohk fine

[muted elbow]

repel is the correct word

[forest token]

👍

[muted elbow]

now since q1 is negative and q2 is positive

do you think there will be a point in between q1 and q2 that will have 0 electric field

[forest token]

no

[muted elbow]

and since in the graph there is no point where E field is zero towards the left of q1

that means q1 must have higher magnitude

cause q1 field is towards right and q2 field is towards left in the region left of q1

[forest token]

oh ok yeah

i have learnt doing this from questions also..find pt of equilibrium of third charge when a system of 2 charges is present

[muted elbow]

yes

the equilibrium point if two charges are of opposite signs always lies outside the charge with smaller magnitude

[forest token]

yeah

ok thanks

.close

[compact pewter]

Channel closed

Closed by @forest token

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[warped oar]

Solve the quadratic congruence:
x^2 = 1 (mod 77)

[compact pewter]

@warped oar Has your question been resolved?

[warped oar]

,w (43^2 - 1)/77

[boreal girder]

Results provided by WolframAlpha

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

Requested by forlorn1002

[warped oar]

I got two solutions.

43 and 34

More generally, 43 + 77k and 34 + 77k

[formal yarrow]

yeah those and 1 and -1

[warped oar]

How?

Prove there are no more.

.close

[compact pewter]

Channel closed

Closed by @warped oar

Use .reopen if this was a mistake.

[bleak dock]

CRT with x^2 = 1 (mod 7) and x^2 = 1 (mod 11)

ah there's also a theorem that says that x^n = k in modular arithmetic has at most n solutions modulo p

to get an idea of how that would work here, suppose that $m^2 \equiv n^2 \pmod p$

[boreal girder]

south

[bleak dock]

then (m - n)(m + n) = 0 mod p, so either m = n or m = -n

modulo p ofc

[bleak dock]

so CRT implies a maximum of 2 * 2 = 4 solutions here

all are possible if you check

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[rough coyote]

can someone help me with this problem please

[rough coyote]

I don't quite understand something about the rotations

.close

[compact pewter]

Channel closed

Closed by @rough coyote

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[silent wigeon]

how is the diameter not x

[idle mountain]

what is a diameter?

[silent wigeon]

a line in a circle

[idle mountain]

a line that PASSES through the center of the circle

[subtle helm]

which?

[silent wigeon]

Oh shit i forgot that it had to pass through the center

[subtle helm]

a line passing through a circle is just a chord (connecting two edges)

[silent wigeon]

what abt radius then? does it have to be half at the centre too?

like half a line before center

[idle mountain]

it is a line that touches the circle and its center

basically half of diameter

[silent wigeon]

Oh alr, if a line doesn't pass through centre. It's not considered a diameter or radius right?

[idle mountain]

yes

[silent wigeon]

Ok thank you

[idle mountain]

it is considered as a chord

[silent wigeon]

👍

.close

[compact pewter]

Channel closed

Closed by @silent wigeon

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[forest token]

how to do this without manually calculating

options btw

[versed sigil]

well, use {x} = x - floor(x), perhaps? 😄

[forest token]

oh right

[versed sigil]

but before that, this expression can be simplified quite a lot ! nevermind!

[forest token]

oh how

[forest token]

im getting a kind of ugly expression

[kind elm]

x - floor(x) is probably your best bet

it shouldn't be too bad

[forest token]

im getting
sqrt(3) - 2(sqrt 2) + 1/(-2 sqrt(3) + 4(sqrt 2)-2)

oh

it cancels

nevermind

[kind elm]

lol

[forest token]

thanks

.close

[compact pewter]

Channel closed

Closed by @forest token

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[spiral jackal]

Could someone please do part ci and cii and see what answer they get

<@&286206848099549185>

[elfin silo]

you want to check if your answer is correct?

[spiral jackal]

i just want someone to tell me the answers for both i and ii

since i think im doing it wrong

just a bit of doubt

[elfin silo]

you can show your work and we can comment if there's any mistake

[spiral jackal]

i did for part c i)

u=(0.27+u)(u+0.2)

[elfin silo]

explain your reasoning so we can see what's wrong

[spiral jackal]

if independent P(J and K) = P(J) x P(K)

P(J and K) = u

P(J) = 0.27 + u

P(K) = (0.2 + u)

[elfin silo]

are you sure P(K) = 0.2+u?

[spiral jackal]

yh thats the part not 100% sure

[elfin silo]

it seems to me L is part of K and 0.2 is only the probabily of K \ (L∪J)

u is that of K∩J

you're missing that of K∩L = L

[spiral jackal]

wait bc the circle is in K does it mean its part of K

[elfin silo]

That's usually how Venn diagrams are read

if you draw a set inside another, it is contained

[spiral jackal]

so what is the probability of K

[elfin silo]

K can be broken into 3 mutually exclusive events
K = (K∩J) ∪ (K∩L) ∪ (K \ (J∪L))

so P(K) is the sum of the probabilities of these events

what are they?

[spiral jackal]

whats (K∩L)

they dont overlap

[elfin silo]

they do, as L sits in K

K∩L = L

[spiral jackal]

(K∩L) = v?

[elfin silo]

P(K∩L) = v

K∩L is an event, its probability is v

[spiral jackal]

btw for c i) it wants it in terms of u

[elfin silo]

sure, let's write it down first though

[spiral jackal]

so is the probability u + 0.27 + v + v

[elfin silo]

there's only one v

no wait, what is that the probabily of?

it should be u+v+0.2

K = (K∩J) ∪ (K∩L) ∪ (K \ (J∪L))
P(K) = P(K∩J) +P (K∩L) +P (K \ (J∪L)) = u+v+0.2

don't you agree?

[spiral jackal]

yeah

so (0.27 + u) x (u+v+0.2)

u+v we told in part b is 0.35

[elfin silo]

yes

[spiral jackal]

(0.27 + u) (0.55)=u

[elfin silo]

there's the equation
if you solve it you find u, with which you find v

[spiral jackal]

yep tysm

for the help

[elfin silo]

you're welcome

[spiral jackal]

btw

[spiral jackal]

are these called subsets?

so L is a subset

[elfin silo]

L is a subset of K, yes

you need to specify what of

[spiral jackal]

makes sense

[elfin silo]

$$L \subset K$$

[boreal girder]

Luigi

[elfin silo]

you write it like this

[spiral jackal]

alr ty

.close

[compact pewter]

Channel closed

Closed by @spiral jackal

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[vapid yacht]

I have done the first part of this question, that is, when a = 1
obviously, a solution only exists if b = c = d

for the second case, that is a not equal to 1, I did gaussian row reduction, in which I fond the following combination of elements in the last row:

$(0 | 0 | \frac{a^4 + 2a - 3a^2}{a^3 - a} | d - \frac{ab + ca}{a(a + 1)})$

actually its best I type out the entire augmented matrix

[boreal girder]

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

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

[vapid yacht]

and then finally

[boreal girder]

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

[vapid yacht]

in this as well, I think there would be two solutions: one with infiintely many solutions and one with a unique solution in which (a - 1) + \frac{(a - 1)^2}{a^2 -1} cannot equal to zero

am I on the right path?

<@&286206848099549185>

[compact pewter]

@vapid yacht Has your question been resolved?

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[forest token]

can someone help me with this? im trying to do it graphically but i dont know how to start

[lime juniper]

call f(x) y and divide the image of f (also known as the range of f, depending on your book) into intervals. have you heard of either term before? i can explain it

[forest token]

yeah i know what range is 💀

could u explain a bit more?

[lime juniper]

yeah sure

[forest token]

about how to divide it into intervals

[lime juniper]

im gonna call it the image because thats what im used to, one sec

[proud halo]

Why don't you do it normally instead of with a graph

You'll need to do the interval dividing anyways

[lime juniper]

[
f(y) =
\begin{cases}
y + 1 & \text{if } y \leq 1 \
5 - y^2 & \text{if } y > 1
\end{cases}
\qquad
g(x) =
\begin{cases}
x & \text{if } x \leq 1 \
2 - x & \text{if } x > 1
\end{cases}
]

notice I renamed g(x) to g(y) but that doesn't matter, you agree?

[forest token]

like our teacher taught a graphical method and i couldnt understand it properly thats why

[forest token]

yeah

[lime juniper]

so, okay, you want to do a graph of this

can I use a computer for it

[forest token]

f(g(x)) =
g(x)+1, g(x)<=1
5-g(x)^2, g(x)>1

[forest token]

yeah

[lime juniper]

oh I confused which was was which , one sec

[boreal girder]

gfauxpas

[forest token]

ohk

[lime juniper]

okay so, the vertical line at x=1 is not actually a vertical line it's a limitation of the program, did you learn that about computer graphs?

[proud halo]

just work with this

solve the inequalities

and you're done

[lime juniper]

yeah but they want a graphical interpretation too which is fine

[proud halo]

seems unnecessary

but sure, you do you

[forest token]

for this question yes but im just using this as a base to learn another method..but in an exam yeah ill solve it using normal method only

[proud halo]

okay

[forest token]

yeah i got it

[forest token]

so how can i graph f(g(x)) from here?

[lime juniper]

alright. what's the image of f?

[forest token]

[-4,4]

[lime juniper]

good

funny stuff happens at x=1, so let's see which half of the curve the points there are on

what is (1,f(1))?

[forest token]

f(1)=2

[lime juniper]

great

notation i'm gonna use:
f([interval]) <- the image of the function f as applied to the interval

f([-2,4)) ? f({4})? f((4,3])?

if you dont understand the notation, I mean that:
f([-2,4)) = [-1,2)

[forest token]

oh ok yeah im a bit unfamiliar with the notation

f of an interval

but yeah i understood

[lime juniper]

ah wait

i was looking at the wrong numbers sotrry

I meant

f([-2,1)) = [-1,2)

[forest token]

f(4) = 5-(4^2)?

[lime juniper]

f({1})={2}
f((1,3]) = [-4,4)

I didnt mean f({4}) that was a mistake

basically this notation means, whats the SET of all output numbers when f is applied to every number in the SET of inputs

[forest token]

👍

[lime juniper]

now repace y with g(x)

[forest token]

yeah

f(g(x)) =
g(x)+1, g(x)<=1
5-g(x)^2, g(x)>1

[lime juniper]

so yeah I actually should have analyed g(x) not f(x)

that was my mistake, if you want to do it graphically

[forest token]

oh

[lime juniper]

yeah it really is hard to do this graphically because doing it with algebra is so tempting, it's only really going to be worth doing the graph if you dont have the formula for the function

[forest token]

😔

[lime juniper]

but sure let's take a look

[forest token]

so we analyse this from x<=1 and x>1 right

[lime juniper]

g([-1,1]) = [-1,1]
g([1,3])=[1,3])
hey that's pretty neat

[forest token]

yeah lol

[lime juniper]

so for f(y), you have:
let y=g(x)
y+1, when g(x)<=1
5-y^2, when g(x) > 1

when is g(x) > 1?

I made a bad typo

[forest token]

uh never?

[lime juniper]

g([1,3])=[-1,1] I should have said
meaning g never exceeds 1 and never goes below -1

[lime juniper]

yup!

so just ignore the 5-y^2 part

[forest token]

oh okay

yeah

and g(x)<=1 for all real numbers

so we just have to graph
x+1,x<=1
and (2-x)+1,x>1

[gloomy sparrow]

y r u trying to graph

[forest token]

bro what is ur problem if i try to graph 😭 i know there are better methods

[lime juniper]

We don't have to graph, Yoda just wants to see how you do it with a graphj

[lime juniper]

this is a different person than before

[gloomy sparrow]

oh okay

[forest token]

yeah i know

that guys my friend lol

[lime juniper]

oh lol

[gloomy sparrow]

i was just wondering y

not complaining

[lime juniper]

anyway a teacher might ask you to do a question like this and not give you the definition of the functions themselves

[forest token]

yeah no i was just learning how to do it using a graph thats it

[lime juniper]

anyawy youre done

[forest token]

oh ok

thanks a lot

.close

[compact pewter]

Channel closed

Closed by @forest token

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[dusk coral]

where is this *3 coming from like i dont get it

[kind crane]

looks like it's the binomial expansion theorem / formula

or are you asking why $\binom{3}{2} = 3$ ?

[boreal girder]

riemann

[dusk coral]

yes

[kind crane]

did you learn $\binom{n}{k} = \frac{n!}{k! (n-k)!}$

[boreal girder]

riemann

[dusk coral]

ahhhh with this formular i get 3

nope didnt know there is this formular

[kind crane]

yea it's important to learn the binomial coefficient to understand what P means in here

[dusk coral]

well thanks a lot

ye true without im doing no progress

well thanks i will close

.close

[compact pewter]

Channel closed

Closed by @dusk coral

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[dapper fable]

is this the right way?

if yes, how to end it?

[compact pewter]

@dapper fable Has your question been resolved?

[compact pewter]

@dapper fable Has your question been resolved?

[compact pewter]

@dapper fable Has your question been resolved?

[kind crane]

you should be using the fact that f is k-times differentiable somewhere since problem 5 suggests Dn * f doesn't need to converge uniformly to f

[dapper fable]

😭😭

Maybe Taylor?

[kind crane]

don't see how that'd help, but sure try it

differentiable -> can do integration by parts

you should also be using fourier representation of f since the question tells you f is periodic

[dapper fable]

Oh god, there is so much things to do haha

But okay, I’ll try it tomorrow and hope it’ll be enough

Do you have any idea about an example for 5?

[compact pewter]

@dapper fable Has your question been resolved?

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[stone jackal]

\subsection*{19) $\begin{aligned}
&h(x, y) = \ln(\sqrt{x^2 + y^2}) \
&(x, y, z) = (3, 4, \ln(5))
\end{aligned}$}
$$z = \ln(\sqrt{x^2 + y^2}) \implies f(x, y, z) = \ln(\sqrt{x^2 + y^2}) - z$$
$$\begin{aligned}
\vec n &= \nabla h \
&= \left<\frac{2x}{x^2 + y^2}, \frac{2y}{x^2 + y^2}, -1 \right> \
&= \left<\frac{6}{25}, \frac{8}{25}, -1 \right>
\end{aligned}$$
$$\frac{6}{25}(x - 3) + \frac{8}{25}(y - 4) - (z - \ln(5)) = 0$$

[boreal girder]

@stone jackal

[stone jackal]

I'm trying to find the tangent plane to h(x, y) at the point (x, y, z) = (3, 4, ln(5)). Where did I go wrong?

.close

[compact pewter]

Channel closed

Closed by @stone jackal

Use .reopen if this was a mistake.

[stone jackal]

,w partial z = ln(sqrt(x^2 + y^2))

[boreal girder]

Results provided by WolframAlpha

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

Requested by en.ratham

[stone jackal]

.reopen

[compact pewter]

✅

[stone jackal]

\subsection*{19) $\begin{aligned}
&h(x, y) = \ln(\sqrt{x^2 + y^2}) \
&(x, y, z) = (3, 4, \ln(5))
\end{aligned}$}
$$z = \ln(\sqrt{x^2 + y^2}) \implies f(x, y, z) = \ln(\sqrt{x^2 + y^2}) - z$$
$$\begin{aligned}
\vec n &= \nabla h \
&= \left<\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{x^2 + y^2}, -1 \right> \
&= \left<\frac{3}{25}, \frac{4}{25}, -1 \right>
\end{aligned}$$
$$\frac{3}{25}(x - 3) + \frac{4}{25}(y - 4) - (z - \ln(5)) = 0$$

[boreal girder]

@stone jackal

[stone jackal]

On Desmos, this looks right to me. Can anyone confirm?

[keen granite]

math appears to check out

[stone jackal]

Thanks!

.close

[compact pewter]

Channel closed

Closed by @stone jackal

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[cyan trench]

I understand that the 1/3 got moved using a rule

But how did that last chunk just turn from positive to negative

[tulip ivy]

Distribution

[jagged sparrow]

It gets transformed to exponent. And 1/3 means cube root.

1/3 as exponent means cube root

cube root of 27 is 3

[tulip ivy]

You mean that, right?

[jagged sparrow]

the last one turned from positive to negative because - x + = -

[cyan trench]

😭

How I miss this

Sorry guys

[jagged sparrow]

Am I correct?

NP

[cyan trench]

Lemme continue tho

[jagged sparrow]

T-T

[cyan trench]

I'll probably run into another issue soon

[jagged sparrow]

GL

[compact pewter]

@cyan trench Has your question been resolved?

Channel closed

Closed by @cyan trench

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[cyan trench]

So I was suppost to simply that

Where does all of the stuff on the right side come from

And how does x need to be greater then 3 but also less then-3

That's impossible

[subtle helm]

Or

It’s or not and

[cyan trench]

Okay

And is this required just cuz there is a denominator at some point,m

[subtle helm]

It’s required from x^2 > 9

[cyan trench]

Why do we need to talk about that

Like what makes us need to do that

Some questions don't have these weird redirections

[subtle helm]

Cuz the input of log

Can only > 0

That’s the domain of log

[cyan trench]

Ok

[cyan trench]

How come my teacher didn't write all of that type of stuff up for this question

[subtle helm]

Weird..

[cyan trench]

But the numerator shouldn't matter right?

Just the denominator

Isek

[subtle helm]

Yes, for this one

From the denominator it’s x > -3

From the initial expression we have x > 1

For it to make sense

[cyan trench]

But for this why do we care about x^2-9

It's not in the denominator

[cyan trench]

So if it's ever in brackets besdie the log?

[jagged sparrow]

Idk where this is going?

What was your question bud?

[cyan trench]

Idk how to explain he is helping tho

I got another one in a sec

[jagged sparrow]

Oh u asked why x<-3 and x>3?

[cyan trench]

Yea got that figure out tho

[subtle helm]

If it’s the input

It must x>0

[cyan trench]

Okay

But I only do that with brackets

Like the brackets ring a bell to check, and besdies that it's x>0

[subtle helm]

Yes u need to check the domain of log

[cyan trench]

If I had this without the brackets then would I have to write x>-5

Or x can't be -5

Or wtf you do

[bleak dock]

so that would become $y = (\log_3 x) + 5$

[boreal girder]

south

[bleak dock]

now +5 is a vertical shift and that doesn't change the position of the vertical asymptote

[bleak dock]

so the domain of that is just x > 0

[compact pewter]

@cyan trench Has your question been resolved?

[cyan trench]

Ok

.close

[compact pewter]

Channel closed

Closed by @cyan trench

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[small brook]

how does this part become the next part?
-how does adding a 1/2 into the integral allow you to integrate?

[stark marsh]

For u sub

[small brook]

.close

[compact pewter]

Channel closed

Closed by @small brook

Use .reopen if this was a mistake.

[small brook]

thx

[stark marsh]

Note that they also multiplied 2 so they didn’t really change the input

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[woven panther]

Number Theory Problem - need hint
https://codeforces.com/problemset/problem/1732/A

Can someone give a hint ?

I tried to apply the operation greedily from the end to the front of the array but I found that it fails.
Then I thought that if an operation is applied on an index and it removes some primes from the element then it should remove all the powers of that prime.
Now for each prime in gcd of all array numbers I should calculate the least cost and if it is possible to entirely remove instances of more than one prime from a number then cost is counted only once.
But I think I have overcomplicated it.

[formal yarrow]

oh wait this one

[woven panther]

yeah, I still didn't get it. I tried something but it got complicated

[formal yarrow]

hint : take ANY array and apply the operation on last two elements, what happens to gcd of whole array now?

[woven panther]

let me try

[formal yarrow]

or better : what happens to gcd of last two elements after the operation

[woven panther]

is gcd(n, n-1) = 1 helpful here ?

[formal yarrow]

yess

exactly

[woven panther]

oooh, let me think

[formal yarrow]

(I explained very badly lol that time)

[compact pewter]

@woven panther Has your question been resolved?

[woven panther]

np

don't give me sol rn, I almost got it

[formal yarrow]

[woven panther]

I think I have formaly proved it

g1 = gcd(n, an) | n
g2 = gcd(n - 1, a_n-1) | n -1

gcd(g1, g2) = 1

n = g1 * k1
n - 1 = g2 * k2
where k1, k2 are non zero integers

after taking gcd on last two numbers if take gcd on these individual gcd's then

gcd (g1, g2) = gcd (n/k1 , (n-1)/k2)

since g1, g2 are integers
k1 | n
k2 | n-1

[woven panther]

gcd (n/k1 , (n-1)/k2) = gcd( (nk2)/(k1k2) , ((n-1)k1)/(k1k2))

= 1/(k1k2) * gcd ((n-1) k1 , nk2)
= 1/(k1k2) * gcd(g2k1k2, g1k1k2)
= gcd(g1, g2)
= 1

[formal yarrow]

you tried to prove gcd(g1,g2)=1 right

you have used gcd(g1,g2)=1 to prove that

circular argument

youre right that gcd(g1,g2) is 1

[woven panther]

oh, right

my bad

[formal yarrow]

but for proof, let gcd(g1,g2)=d
from g1 | n, g2 | n-1
you get d | n and d | n-1

continue from here

in simple words, g1 is a divisor of n, g2 is a divisor of n-1 and since n and n-1 don't have any common factors, g1 and g2 also don't have any common factors

[woven panther]

Oh, I finally got it intuitively

thanks

[formal yarrow]

nice

now you can make the algorithm

[woven panther]

yeah, that's the easy part

[formal yarrow]

[woven panther]

accepted

🥳

.close

[compact pewter]

Channel closed

Closed by @woven panther

Use .reopen if this was a mistake.

[formal yarrow]

🎉

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[tropic agate]

I'm reading about condition numbers $\operatorname{cond}(A)$. I'm faced with the true or false question "If $Ax=b$ is well-conditioned, then $\operatorname{cond}(A)$ is small." I think it is false, and I think a counterexample is given by the matrix $$\begin{pmatrix} k^2&0\ 0&1 \end{pmatrix}\text{ with } k\gg 1.$$I'm just learning the ropes of this; how do I show a system with the above matrix is well-conditioned?

[boreal girder]

psie

[compact pewter]

@tropic agate Has your question been resolved?

[hushed magnet]

how did you define well conditioned systems

did you not define it by using that cond(A) is small?

its been some time since I've done that kinda stuff. but wikipedia does define it in that way

[tropic agate]

well, it is true I think that if cond(A) is small, then the system is well-conditioned. But this is the converse. A system is well-conditioned if small relative changes in the coefficients of the system cause small relative errors in the solution. So for example, consider x+y=5 and x-y=1. This system has the solution (x,y)=(3,2). Then consider x+y=5 and x-y=1.00001, which has the solution (3.00005,1.99995). Thus this system is well-conditioned.

[tropic agate]

.close

[compact pewter]

Channel closed

Closed by @tropic agate

Use .reopen if this was a mistake.

[hushed magnet]

the maximum relative error is precisely given by the condition number

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[tropic agate]

I think I found a counterexample. Consider $b=(0,1)$ and also $b'=(0,1+h)$ where $h$ is small. Then the solutions are $x=(0,1)$ and $x'=(0,1+h)$, and the relative change in $b$ is $|b'-b|/|b|=|h|$ whereas the relative change in $x$ is $|x'-x|/|x|=|h|$, so the system is well-conditioned. But $\operatorname{cond}(A)=k$ can be made arbitrarily large.

[boreal girder]

psie

[hushed magnet]

ok but thats not all possible errors

what about b'=(h,1)

or something

[tropic agate]

yeah, you're right. I didn't consider all errors. Do you know any efficient way to assess all errors of the system?

[hushed magnet]

https://i.imgur.com/DdCjceG.png

but I mean your system is 2x2. you could just let the error be e=(h1,h2) and then do all the calculations

[tropic agate]

interesting. How come we are only considering error in b? Shouldn't one consider errors in A as well?

[hushed magnet]

have you not seen that before? how did your course introduce the condition if not with this

not sure why errors in A are not considered

[compact pewter]

@tropic agate Has your question been resolved?

[tropic agate]

I think I finally understand now. We have if $A$ is invertible, that $$\frac1{\operatorname{cond}(A)}\Delta b\leq \Delta x\leq \operatorname{cond}(A)\Delta b,$$where $\Delta b$ and $\Delta x$ are the relative changes in $b$ and $x$. So if the system is well conditioned, we have $\Delta b\approx\Delta x$, so divide the above inequality by $\Delta b$, then $\Delta x/\Delta b=1+\epsilon$ where $|\epsilon|$ is small. And so the inequality reduces to $$\frac1{\operatorname{cond}(A)}\leq 1+\epsilon\leq \operatorname{cond}(A).$$Clearly this does not imply $\operatorname{cond}(A)$ is small.

[boreal girder]

psie

[tropic agate]

What I used above is the following theorem, @hushed magnet :

.close

[compact pewter]

Channel closed

Closed by @tropic agate

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[worn wyvern]

\begin{align*}
\frac{e^{2x} - 1}{6e^x} &= \frac{2e^{2x} \cdot 6e^x - 6e^x (2e^{2x} - 1}{6^2 e^{2x}} \
&=\frac{2e^{2x}-6e^x\cdot e^{2x} +6e^x}{6^2e^{2x}}
&=
\end{align*}

[worn wyvern]

pressed enter by mistake

[keen granite]

https://tenor.com/view/crying-emoji-dies-gif-21956120

[fluid stag]

this is what you have written

[boreal girder]

<rajel />

[worn wyvern]

.close

[compact pewter]

Channel closed

Closed by @worn wyvern

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[carmine garden]

can I have a hint

[carmine garden]

I was thinking of starting by saying v \in U and U is spanned by e_1,e_2,\dots,e_m

So $T( \sum_{i=1}^{m} \alpha_i e_i)= \sum_{i=1}^{m} \beta_i e_i$

[boreal girder]

wai

[slate sand]

What does the invariant part mean, T(U) = U for every T:V->V linear?

[carmine garden]

the domain and co-domain are the same

[slate sand]

So T(U) = U

[carmine garden]

yes

[slate sand]

Is this question under dual spaces?

[carmine garden]

no

[slate sand]

Before?

[carmine garden]

after

I need dual spaces here? Denascite suggested I come dual spaces to this later

[slate sand]

I don’t know I’m thinking

[carmine garden]

I was thinkng contrapositive

consider a vector space that's neither V not {0} . Then there exists a linear transfomration that's not invarient

[slate sand]

I don’t think this is right because how can this be true

[carmine garden]

wdym

[slate sand]

Codomain or range/image

[carmine garden]

image

my bad, yes

[slate sand]

I mean

Is it even true that every operation is invariant on V itself

The map that sends everything to 0 maps V to 0 not V to V

Or does it just mean that T(U) ⊆ U

[carmine garden]

Yea, this

The image is still in V

[hybrid walrus]

I find that taking a basis of U and expanding it to a basis of V makes this very simple

[slate sand]

Okay this makes a lot more sense then

That would be my thought too

[carmine garden]

Won't contrapositive work better?

[carmine garden]

How would this work?

[hybrid walrus]

Contrapositive, yeah. Take the linear operator on V that swaps a basis element of U with a basis element not in U (which both exist by assumption).

[carmine garden]

Yea, that was my thought too for contrapositive

I thought you were talking about direct proof

😭

thanks

.close

[compact pewter]

Channel closed

Closed by @carmine garden

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[deft tangle]

I didn't understand the question properly what they are giving and asking

Information is too much tricky

[carmine garden]

Given:$p,q \in \Z$ . $\alpha, \beta$ are the roots of $x^2-x-1$. $a_m= p \alpha^n + q \beta^n$. If $a,b \in \Q$ and $a+ \sqrt{5}b=0 \implies a=b=0$.
\
If $a_4=28$, find $p+2q$

[boreal girder]

wai

[kind viper]

hang on

alpha and beta are the roots of x^2 - x - 1 = 0 sure

but do we know which is which?

i think the question is wrong

[west canyon]

it could be a special case like p=q in this instance so it doesn't matter

[kind viper]

i mean ok like p alpha^4 + q beta^4 = 28

[west canyon]

wait yeah it will be this cus ||if you let \sigma be the other element of Gal(x^2-x-1), then applying sigma to both sides should give you what you need||

[kind viper]

that's quite the nuke LMAO

but i did also work it out the honest way

[west canyon]

fair lmao

[kind viper]

well it has the benefit of respecting nosols

[compact pewter]

@deft tangle Has your question been resolved?

[short ferry]

notice that the equation can be rearranged into x^2 = x + 1

both α and β are roots of that equation, that means
α^2 = α + 1
β^2 = β + 1

try using this to reduce the powers in a_4

[kind viper]

ie try expressing alpha^4 as linear function of alpha

[gleaming field]

What is the square root of 1690

[kind viper]

!occupied

[compact pewter]

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).

[short ferry]

around 41.1, if it's not related to the question though, move to another channel. If it is how you are solving the question, then you are likely approaching it in a wrong way

[gleaming field]

Alr mb

[compact pewter]

@deft tangle Has your question been resolved?

[deft tangle]

a_4=28

=Pa^4+qB^4

[short ferry]

okay yeah

so 28 = Pa^4+qB^4

[deft tangle]

Yeah

[short ferry]

now use this

btw you can multiply it with a^2

[deft tangle]

α^2 = α + 1
β^2 = β + 1

[short ferry]

so you get a^4 = a^3 + a^2

[short ferry]

now you can substitute that inside this

and you repeat it for a^3 and a^2

[deft tangle]

i did not understand this

Ohh i got it

Yes

[short ferry]

yeah, so just keep reducing this thing

[deft tangle]

so 28 = Pa^3+Pa^2+qB^3+qB^2

@short ferry

What should I do next?

[short ferry]

Continue

rewrite a^3 using a^2 + a

then a^2 using a and 1

and same for beta

[deft tangle]

28=2Pa^2+P+2qB^2+q

@short ferry

[short ferry]

hmm

that doesnt look right

shouldnt it be Pa and qB

28 = 2pa^2 + pa + 2qb^2 + qb

do you know fibonacci btw?

notice that a^n acts kinda like it

[deft tangle]

Nope

[short ferry]

okay, nvm then

just continue doing it

until you reach something so simple, that it has only a and b in it with no powers at all

[compact pewter]

@deft tangle Has your question been resolved?

[compact pewter]

@deft tangle Has your question been resolved?

[compact pewter]

@deft tangle Has your question been resolved?

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[cedar mortar]

How do I calculate the test statistic and the p-value by using the TI-84 Plus CD calculator? I need help!

[rigid perch]

stat -> tests -> 6: 2-prop Z-test

[compact pewter]

@cedar mortar Has your question been resolved?

[cedar mortar]

so what numbers do put?

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[rigid perch]

. n = sample size, x = number who say "yes" = n * p

[cedar mortar]

Thanks!

[cedar mortar]

.close

[compact pewter]

Channel closed

Closed by @cedar mortar

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[digital wraith]

Can you guys help me with this question:

When a polynomial is divided by x-1 the remainder is 3. When the polynomial is divided by x+2 the remainder is -12. Find the remainder when the polynomial is divided by x^2+x-2

[digital wraith]

idk where to start

[kind viper]

ok right

When a polynomial is divided by x-1, the remainder is 3.

call your polynomial P(x).
do you know how you could rephrase this sentence?

[digital wraith]

no I didn't really listen when my teacher was going trhough a question similar to this

[kind viper]

mmmmkay

ok

[digital wraith]

soz

[kind viper]

do the words "remainder theorem" sound familiar?

[digital wraith]

yeah

i have some notes for it

[kind viper]

ok can you state the remainder theorem

[digital wraith]

its like a proof tho

Suppose that P(x) is a polynomial and α is a constant. Prove that the remainder after dividing P(x) by (x-α) is P(α)

P(x) = (x-α) Q(x) + R(x)

P(α) = (α-α) Q(α) + k

[kind viper]

well ok it would be good to distinguish a theorem and its proof

[digital wraith]

ok

[kind viper]

the remainder after dividing P(x) by (x-α) is P(α)
this is what you need

you prove it once -- presumably, you already did in class one way or another -- and then you use it later.

[digital wraith]

so do i use that to solve my question

[kind viper]

yes

When a polynomial is divided by x-1, the remainder is 3.
use the remainder theorem to rephrase this

as P(__) = __

fill in the blanks

[digital wraith]

do i jsut sub in x-1

[kind viper]

i am extremely hesitant to instruct you to just "sub in" anything into anything.

[digital wraith]

oh

sorry im year 10 doing accelerated im a bit dum

[kind viper]

general theorem:

When P(x) is divided by (x-a), the remainder is P(a).
your question says:
When P(x) is divided by (x-1), the remainder is 3.

fill in the blanks in P(__) = __

[digital wraith]

P(1) =

[kind viper]

you filled in one blank, now fill in the other.

what number is P(1) equal to?

[digital wraith]

3?

[kind viper]

yes

so P(1) = 3.

good.

When the polynomial is divided by x+2 the remainder is -12.
now do the same here. rephrase this statement using the remainder theorem.

exact same logic as i just walked you through, only the numbers are different.

[digital wraith]

P(-2) = -12

?

[earnest wyvern]

hey did this type of questions in my preparation for additional mathematics

wait i might help maybe

[digital wraith]

so if P(1) = 3 and P(-2) = -12 how do i find p(x+2)(x-1)

[earnest wyvern]

i m trying

just give me 2 mins

ill tell u

[digital wraith]

ok thank you rajput

[earnest wyvern]

actually very interesting question

[kind viper]

not the right notation

[boreal nimbus]

the divisor (x + 2)(x - 1) is a quadratic, so the remainder is of form Ax + B

[earnest wyvern]

yes

i was gonna say that

[kind viper]

!nosols

[compact pewter]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[kind viper]

(preemptive)

(but just as a reminder)

[earnest wyvern]

uh.....

i know the answer

i have solved it

[digital wraith]

idk what to do

im still very confused

[earnest wyvern]

can i tell u

-36 is the remainder

[digital wraith]

can u explain the

theorem

[earnest wyvern]

ok

[digital wraith]

i dont get it

like how to apply it

so i can actually do other similar questions bc i dont understand how to do these questions

but sketching polynomials is fine for me

[kind viper]

!nosols

[compact pewter]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[kind viper]

@digital wraith ok so let's pick this back up shall we

[digital wraith]

ok

please

[kind viper]

you know that P(1) = 3 and P(-2) = -12.

[earnest wyvern]

when p(1) = 3 and p(-2) = -12 and after factorization of x^2 +x -2

i m writing sol

[digital wraith]

i have to get my homework done sicne i have exam block to worry about

[kind viper]

with all due respect please back off i've already reminded you of server policy twice here

dont dump the entire solution at him

[earnest wyvern]

ok sorry

i didnt understood, my bad

[kind viper]

the division-with-remainder of P(x) by (x^2 + x - 2) will look like this:

P(x) = (x^2 + x - 2)Q(x) + Ax + B

since x^2 + x - 2 is a polynomial of degree 2, any remainder on division by it will look like a linear or maybe constant polynomial

[earnest wyvern]

wont happen again

[kind viper]

@digital wraith do you understand this?

[digital wraith]

i understand the 1st part

cuz it was the wroking

[earnest wyvern]

why r u telling it then

[digital wraith]

for the example we did in class

but then i dont understand the remainder on divison part (last part)

[earnest wyvern]

do u see a connection

[kind viper]

i am taking him through it step by step mate

im not going to dump the whole thing at him

i already asked you to back off

i do not want to repeat myself here

you're kind of interrupting and getting in my way and honestly i am a little frustrated at that

[earnest wyvern]

(x-1) has remainder 3 and x+2 has remainder -12

[boreal nimbus]

(this is wrong, by the way)

[kind viper]

<@&268886789983436800>

[digital wraith]

the answers are at the back of my book and it says 5x-2

im not gonna copy

it cuz i wanna understand steps

to do good in quiz

its for tutoring

[kind viper]

im trying to explain it but rajput is getting in the way here and it is hard for me to get a word in.

ok

[digital wraith]

ok

[kind viper]

let me repeat my last phrase

since x^2 + x - 2 is a polynomial of degree 2, any remainder on division by it will look like a linear or maybe constant polynomial

do you understand this part

like, the remainder of a polynomial division always has strictly lower degree than the divisor

for a divisor whose degree is 2, this means remainder has degree 1 or 0

[digital wraith]

oh yeah

[earnest wyvern]

so (x-1)(x+2) will be remainder of first equation i.e 3 multiplied with remainder of second equation which is -12. 3 x (-12)

[digital wraith]

i get it now

for that part

since something squared

[kind viper]

this is just plain wrong lmfao

[digital wraith]

divided by like x-1 or smth will give

soemthing to the power of 1

👍

[earnest wyvern]

how

[digital wraith]

i get it

[kind viper]

divising by x-1 or x-a in general will give a constant (and you know precisely what constant)

[digital wraith]

ye its 3 for when P(x) when divide by x-1

[kind viper]

x+10 divided by x gives remainder 10 and divided by x-1 gives remainder 11 but x+10 divided by x(x-1) does not give remainder 110.

anyway

[earnest wyvern]

no

it doesnt

i m not saying it gives 110

[kind viper]

@earnest wyvern please take this argument to your own help channeol

[digital wraith]

i just wanna learn

[kind viper]

P(x) = (x^2 + x - 2)Q(x) + Ax + B
are we good on this

[digital wraith]

yes its from the formula

[kind viper]

ok right

now you know P(1)

so put x=1 into this

dont simplify yet

ie leave P(1) as P(1) for the time being

[earnest wyvern]

d4n atleast read my messages maybe you will understand.

dont get pissed i m going