help-43

[deft tangle]

Yeah

Opps sorry

2π for sinx

Yes for mod it is π

[lone quartz]

Right

[deft tangle]

|sin2x| will be π/2

[lone quartz]

Yup

[deft tangle]

|sin4x| will be π/4

[lone quartz]

Yes good

[deft tangle]

Now?

[lone quartz]

Now you try to find it for cos4x in the similar way

[deft tangle]

Same

[lone quartz]

Yes

[deft tangle]

I guess answer should be π/8

[lone quartz]

No

If two functions of same period are added then the period of resulting function is also same

[deft tangle]

I don't think

[lone quartz]

I think it's correct. Let me check

Yh mb

It's half of it

[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.

[deft tangle]

Can anyone explain these points how they find order of it

[copper sierra]

@deft tangle list the elements and compute their powers

[deft tangle]

Could you help with D5?

Let me search over google

[copper sierra]

do u have trouble computing powers?

[compact pewter]

@deft tangle Has your question been resolved?

[compact pewter]

@deft tangle Has your question been resolved?

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.

[vivid siren]

number of discontinuity of 1/log|x|

[vivid siren]

so at x=1,-1 will not be there

[chilly basalt]

log(x), what’s the restriction of x?

[carmine garden]

what is the domain of log(x)?

[compact pewter]

@vivid siren Has your question been resolved?

[vivid siren]

x=1,-1

[vivid siren]

x=-1,-1 rest all R

[chilly basalt]

Nope, you clearly miss something

And -1 is an illegal term to plug in log(x)

[compact pewter]

@vivid siren Has your question been resolved?

[vivid siren]

we have mod

[vivid siren]

so

logx can not take negative numbers but log|x| can take

[chilly basalt]

Correct, what else is an illegal term to plug in log(x)?

[deft tangle]

Log0?

[chilly basalt]

Correct, but I expect OP to respond instead

[deft tangle]

Ahh sry my bar

[compact pewter]

@vivid siren Has your question been resolved?

[compact pewter]

Channel closed

Closed by @vivid siren

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.

[vivid siren]

hints please

[young raft]

if it passes through the origin, plugging in x=y=z=0 should help you

[vivid siren]

yes

c=0

[vivid siren]

hey

[azure vault]

wait no ok

[vivid siren]

?

[azure vault]

nono don't worry

now that we found c = 0

[vivid siren]

yeah

[azure vault]

it's good to remind ourselves that you can write the equation of a sphere as:

(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = r^2

where x_0,y_0,z_0 and r are all constant

(x_0,y_0,z_0) is the center of the sphere

r is its radius

[vivid siren]

yeah

[azure vault]

now comparing this form

[azure vault]

with this one

[vivid siren]

i see

[azure vault]

doesn't some term feel... out of place?

[azure vault]

something that you would never see even if you expanded this?

[vivid siren]

x_0^2+y_0^2+z_0^2=r^2

[young raft]

yeah it's weird

[azure vault]

that's if the center is (0,0,0)

wait

no ok

I get what you did

you're trying to find a link between the coordinates of the center and the radius

but we know neither

so

[azure vault]

expand this

see if there's anything from the other equation that would never match

and deduce what it implies on a,b,c

(well we already know c = 0)

[vivid siren]

( x^2 + x_0^2 + 2xx_0 + y^2 + y_0^2 + 2yy_0 + z^2 + z_0^2 + 2zz_0 = r^2 )

[boreal girder]

Tom

[azure vault]

with minuses

[vivid siren]

ohh yes

$$
x^2 + x_0^2 - 2xx_0 + y^2 + y_0^2 - 2yy_0 + z^2 + z_0^2 - 2zz_0 = r^2
$$

[boreal girder]

Tom

[azure vault]

ok, and to be simpler

$x^2 - 2xx_0 + y^2 - 2yy_0 + z^2 -2zz_0 = \text{some constant}$

[boreal girder]

rafilou is not not born in 2003

[azure vault]

now, with that said

[vivid siren]

you said

[azure vault]

?

yes

[vivid siren]

x=y=z=0

$$
x_0^2 - 2xx_0 + y_0^2 - 2yy_0 + 2 - 2zz_0 = r^2
$$

[boreal girder]

Tom

[azure vault]

and... the other x,y,z?

didn't you set them to 0 too?

[vivid siren]

x=y=z=0

[azure vault]

also what happened to z_0^2

$$
x_0^2 - 2xx_0 + y_0^2 - 2yy_0 + z_0^2 - 2zz_0 = r^2
$$

[boreal girder]

rafilou is not not born in 2003

[azure vault]

but with x = y = z = 0

so

[vivid siren]

$$
x_0^2 - 2xx_0 + y_0^2 - 2yy_0 + z_0^2 - 2zz_0 = r^2
$$

[boreal girder]

Tom

[azure vault]

$$
x_0^2 + y_0^2 + z_0^2 = r^2
$$

[boreal girder]

rafilou is not not born in 2003

[vivid siren]

yes

[azure vault]

again

that won't help us

because we know none of those 4 constants

at least that will not help us yet

we have to find the center of the sphere first

[vivid siren]

ok so?

[azure vault]

that's where we start looking back at the original equation

and comparing

[vivid siren]

yes

[azure vault]

$$ax^2 + 2y^2 + 2z^2 + 2bxy + 4x = 0$$
is the same as
$$x^2 - 2xx_0 + y^2 - 2yy_0 + z^2 -2zz_0 = \text{some constant}$$

[boreal girder]

rafilou is not not born in 2003

[azure vault]

so

with that said

we should be able to find a and b

[vivid siren]

i did not get your point

[azure vault]

?

[vivid siren]

one equation we made by general way

one we are given so what we are comparing

we got nothing??

[azure vault]

we are gonna get something

we have a mystery sphere to find

[vivid siren]

a=1

[azure vault]

and we are given its equation

[vivid siren]

b=2

?

so how we will find it?

[azure vault]

If I wrote that 2x + ay = 1

and I knew that x + y = constant

I should be able to find a and the constant, right?

[vivid siren]

2x+2y=2constant

a=2

i need to know the constant value

[azure vault]

don't you have it now?

2x + 2y = 1

2x + 2y = 2constant?

[vivid siren]

x + y = constant

here?

[azure vault]

and what is the constant equal to?

[vivid siren]

equal to 1

[azure vault]

2x+2y = 1

but at the same time

2x+2y = 2constant

[vivid siren]

i am saying for value of a

[azure vault]

?

we found a = 2

[vivid siren]

no

[azure vault]

so 2x+ay = 1 becomes 2x+2y = 1

[vivid siren]

that is wrong

[azure vault]

you wrote it yourself

[vivid siren]

2x + 2y = 1
2x + 2y = 2constant

you wrote

x+y=constant

[azure vault]

isn't x+y = c the same as 2x+2y = 2c?

i didn't bother giving a name to my constant

if that's what bothered you then sorry

[vivid siren]

2x+ay=1
x+y=constant

2x+2y=2 constant

(2-a)y=2constant-1

the constant will affect the value of a no?

[azure vault]

?

[vivid siren]

that is why i am saying a=2 could be wrong

[azure vault]

no, it's correct

because if a is not 2

then something that varies = something constant

as you wrote with (2-a)y=2constant-1

[vivid siren]

different constant diffferent values

[azure vault]

different constant?

the constant is a constant

it's fixed

[vivid siren]

which you assumed

[azure vault]

it doesn't vary

[vivid siren]

but each constant will vary

[azure vault]

????????????????????

[azure vault]

then it's not a constant?

[vivid siren]

bro i don't know you are making confusing me on a tiny thing

and i am not getting your final point

if you want me to test on solving linear equations

[azure vault]

no, I want to test you on identifying constant coefficients

"the equation of a line is 2x+ay = 1"

"the equation of that same line is "x+y = c"

"find all constants"

you have two equations

that describe the same object

so we should be able, by manipulating both equations

to find out what's missing in each

[vivid siren]

2x+ay = 1
x+y = c

2x+2y=2c
(2-a)y=c-1

you are doing two things

finding the value and showing the lines same

i know the equations of same slope will be same

they will only differ by a constant

[azure vault]

we KNOW the equations make the same line

because it's told to you in the statement

[vivid siren]

but i am saying you that the value of that constant will affect the value of a

[vivid siren]

what it is told?

[azure vault]

I literally created a question statement for you

[vivid siren]

you did not say it earlier

so i said that it will vary

solving equations and saying same line repersentation

[azure vault]

I thought it was pretty implicit

[vivid siren]

okay so what you wanna do it with same

[azure vault]

and when I rewrote it

[vivid siren]

back to my question

i have compared and told you values of a and b

but you denied

[azure vault]

yeah, because your values are incorrect

in the original problem, you said

a = 1, b = 2

let's rewind

$$ax^2 + 2y^2 + 2z^2 + 2bxy + 4x = 0$$
is the same as
$$x^2 - 2xx_0 + y^2 - 2yy_0 + z^2 -2zz_0 = \text{some constant } ( = r^2 - x_0^2 - y_0^2 - z_0^2) = 0$$

[azure vault]

those are two equations

for the same sphere

so now

manipulate both equations

[vivid siren]

how did you find second equation?

[azure vault]

.

[vivid siren]

no

[azure vault]

yes

[vivid siren]

how you removed x_0

x y z varies

[azure vault]

????

[vivid siren]

so if it passing through origin they should removed

[azure vault]

??????????

[azure vault]

I can't understand you

[vivid siren]

same to me

[azure vault]

no, that's for x=y=z=0

there are still x,y,z appearing here

[vivid siren]

i am asking how will you remove

x_0,y_0,z_0

[boreal girder]

rafilou is not not born in 2003

[vivid siren]

ohh wait

=0?

[azure vault]

well yes

[vivid siren]

and why?

[azure vault]

we found that x_0^2 +y_0^2 + z_0^2 = r^2

by plugging in x=y=z=0

on the sphere

[vivid siren]

yes we found it because it passes through origin

ohh got it

r^2-r^2

you said earier it is useless?

no

[azure vault]

well we found some use to it

but I had in mind that we would reuse it later on

[vivid siren]

do not say it pls

[azure vault]

I didn't think about how it could be useful beforehand

[vivid siren]

it is helping me as you can see

[azure vault]

Yes I was wrong at that time

ok

[azure vault]

$$ax^2 + 2y^2 + 2z^2 + 2bxy + 4x = 0$$
is the same as
$$x^2 - 2xx_0 + y^2 - 2yy_0 + z^2 -2zz_0 =0$$

[boreal girder]

rafilou is not not born in 2003

[vivid siren]

a=1?

[azure vault]

now, to find a,b

[azure vault]

try again

you should think about manipulating one equation or the other

[vivid siren]

hm wait

a=2

[azure vault]

yes

[vivid siren]

b=0

[azure vault]

yes

there are no "xy" terms

[vivid siren]

i see now i got what you were tellling me

thanks sorry by my side

[azure vault]

its alr

ok now

$$2x^2 + 2y^2 + 2z^2 + 4x = 0$$
is the same as
$$x^2 - 2xx_0 + y^2 - 2yy_0 + z^2 -2zz_0 =0$$

[boreal girder]

rafilou is not not born in 2003

[azure vault]

let's go find the center?

[vivid siren]

a=2,c=b=0

[azure vault]

what can we simplify immediately here?

on the first equation

[vivid siren]

x^2+y^2+z^2+2x=0

[azure vault]

I'll let you find x_0,y_0,z_0 now

it should be easy with that

[vivid siren]

y_0,z_0 will be 0

and x will be 1

so radius will be 1

[azure vault]

your final answer is correct, but x_0 is wrong

sign

[vivid siren]

-1

(-1,0,0)

thanks rafilou

[azure vault]

np

[vivid siren]

which book did you read for 3D?

[azure vault]

uuuhhh

none 😅

[vivid siren]

lmao

.close

[compact pewter]

Channel closed

Closed by @vivid siren

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.

[charred parcel]

Right angled triangle, the lines are bisectors, no further info on the subject. I need to find the angle BDA

[charred parcel]

I'm short on time too

[compact pewter]

@charred parcel Has your question been resolved?

[charred parcel]

help

<@&286206848099549185>

[compact pewter]

@charred parcel Has your question been resolved?

[charred parcel]

I still need help

[worldly hatch]

Do you have the ab, bc and ac lengths

[charred parcel]

no

[charred parcel]

screw this

.close

[compact pewter]

Channel closed

Closed by @charred parcel

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.

[quartz yoke]

x^2 = 9^2 + 6^2

[visual musk]

notice how all radii of the circle have the same length, is there a different way you can draw the radius to potentially find it's length?(Purple Relapse gave a good hint)

[neon coral]

I think first is 15

[visual musk]

yup :)

[quartz yoke]

wait wha

oh 12 might be length of half the chord

[visual musk]

yeah, that's why it's moreso to the left than the second problem

[compact pewter]

@neon coral 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.

[wild hill]

i rushed my hw but does this make sense?

[wild hill]

sorry for any bad hand writing i have school in less than 30 it was really rushed 😭

[gentle bear]

Looks good to me

The only thing thats not 100% clear is your justfication for why sinx>cosx for pi/4<x<pi/2 and cosx<sinx for 0<x<pi/4 but whether that's needed in this case might depend on your teacher/curriculum

It is correct though.

[wild hill]

okay that’s great cause my notebooks already packed and everything but i’ll ask my teacher since it’s only high school calculus

and ty!!

.close

[compact pewter]

Channel closed

Closed by @wild hill

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]

if I draw a circle with taking 5i as center then only point which lines on its circumference will be C?

[kind viper]

what does $I(z)$ mean?

[boreal girder]

Ann

[deft tangle]

I don't know either

I guess it is for option ABCD to be correct?

[gentle bear]

Would be best to confirm but I imagine it means the imaginary part of z, particularl since one of the given answer choices is correct with that guess.

[deft tangle]

C

[kind viper]

ok then all speculation is useless

if you do not know the meaning of all notation used in the question then you should not be doing it

[deft tangle]

Hmm i was confused too at the start you are right

Given answer is C

.close

[compact pewter]

Channel closed

Closed by @deft tangle

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]

How can I start this problem? Should I give random numbers like a=b=c=d=0?

A^5-A^3=I

A^3(A^2-1)=I

A^3=I

Or A^2=1

[kind viper]

A^3(A^2-1)=I
A^3=I

Or A^2=1

this is wrong

[deft tangle]

😫 😫

[kind viper]

even for ordinary numbers it would be wrong: from xy = 1 you CANNOT say that x=1 or that y=1

also $A^5 - A^3 = A^3(A^2 - I)$ not $A^3(A^2 - 1)$

[boreal girder]

Ann

[kind viper]

the equation $A^3(A^2 - I) = I$ is still useful though

[deft tangle]

My bad first wrote I

[boreal girder]

Ann

[deft tangle]

So A^3=I

[kind viper]

because you can in fact see that your marked answer "an invertible matrix" is correct

[deft tangle]

A^2=I

[deft tangle]

How can I use it?

A^3 and A^2?

[kind viper]

no you're overthinking it

the statement $A$ is an invertible matrix'' means there exists a matrix $B$ such that $AB = BA = I$''.

[boreal girder]

Ann

[kind viper]

in fact AB = I implies BA = I for matrices, so just one direction is enough

here, you have $A \cdot A^2(A^2-I) = I$

[boreal girder]

Ann

[kind viper]

thus $A^{-1}$ exists.

[boreal girder]

Ann

[deft tangle]

A^2(A^2-I) would be B?

[kind viper]

yes, that's what plays the role of B for us

my point is that we don't care a lot what B is exactly, just that it exists

[deft tangle]

Yeah. That's nice

So should I check option A,B?

Even it was not multiple choice question but

[kind viper]

hmm

i actually don't know if you could extract any info about symmetry or skew-symmetry here.

[deft tangle]

Thanks ma'am ☺️

[kind viper]

i think you couldn't. if a matrix A satisfies A^5 - A^3 = I, then any conjugate matrix to it will satisfy the same property -- but (skew) symmetry is not preserved under those

[compact pewter]

@deft tangle Has your question been resolved?

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.

[obtuse remnant]

How do I know if I’m getting min or max in optimization questions? The process to the solution seems the same unless I’m missing something

1. Find two numbers who’s sum is 60 and their product is a maximum

2. Find two numbers who’s difference is 40 and their product is a minimum

[plain quail]

Use the first derivative test

[obtuse remnant]

I just differentiate both and I find the answer so I’m not sure what the difference would be

Like for 1 how would I find the minimum product instead

[inland nacelle]

AM GM

maybe thats not the method here

[plain quail]

What topic are you guys learning in school

Pretty sure it's a calc qn

[inland nacelle]

oh

[obtuse remnant]

Yeah it’s Calc

[plain quail]

So you solve for the stationary points

Find the nature of the points

And also evaluate the end points

So for the 1st qn

a+b=60

and we're optimizing a•b

So we optimise b•(60-b)=60b-b^2

We find stationary points:
60-2b=0
b=30

So in this case there's only one stationary point

But in some cases there might be two

[obtuse remnant]

Oh ok gotcha thanks so for that question there wouldn’t really be a minimum product

[plain quail]

Yea

[obtuse remnant]

So to figure out if finding the derivative will get me the max or min I have to graph it?

[plain quail]

No is just visualisation

Uh nvm omega is here

[obtuse remnant]

Ooh ok I think I get it now

[thorny urchin]

you can use the second derivative test (which could potentially be inconclusive)
or consider slopes around your location

[plain quail]

The latter of which is the first derivative test

[obtuse remnant]

Gotcha thank u that makes sense now

Just relearnt curve sketching so forgot about how it worked

Thanks everyone

.close

[compact pewter]

Channel closed

Closed by @obtuse remnant

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.

[sly badger]

Does this proof make sense ?

[worldly hatch]

uhm...i'm not sure

how do you make the jump from "smaller than 1 + 2^(-k-1)" to "smaller than 1"

[true pecan]

No

To further elaborate what @worldly hatch already pointed out:

By IH we have s_n < 1, okay. Now s_(n+1) < 1+1/2^(n+1). Still okay, but not helpful. This doesn't mean that s_(n+1) < 1 though and you don't get the the induction hypothesis back.

Another mistake that you did is even if you were showing correctly that the infinite series is less than 1/n+1 for all natural n. This would not mean that the series wouldn't eventually be 1

[sly badger]

I never said that I said the series converges at 1

1 is the limit

When n goes closer to infinty

[true pecan]

Series convergence (in ana 1) is defined to be the limit of the partial sums?

[sly badger]

Ok now this talk is a bit too advanced

[true pecan]

Also you wrote "Since the series converges at 1"

[quaint gulch]

if you just want to prove the statement, you don't need to talk about post highschool math
you could even choose to show, by induction, that the sum is equal to 1-1/2^n, which is smaller than 1

[sly badger]

Sure

Is it because the series increases by 1/2^k each time

[quaint gulch]

yes, 1/2^(n+1) is half of 1/2^n

so you only remove half of the difference with 1 with each term

[sly badger]

Sure so instead of saying that the series is less than 1 I can say that the series is 1-1/2^k

Which means ,

[quaint gulch]

which means it's less than 1

since 1-1/2^k is smaller than 1 without much doubt

[quaint gulch]

yup

[sly badger]

And proven ?

By induction

I guess before I say that the series is 1 - 1/2^k I have to explain why can you tell me how I would say that like fluently so that someone can understand , english is not my first language

[quaint gulch]

you did induction yeah

I mean if you prove by induction you don't need to know why
but a way to prove it is to notice that (1-1/2)(1 + 1/2 + 1/4 + ... + 1/2^n) = 1-1/2^(n+1)

[sly badger]

Wait can I say like since the sum is less than 1 it is equal to 1 - 1/2^k then follow through with the steps I showed you

[quaint gulch]

the steps you showed should come first since saying the sum is less than 1 is what you want to prove

[sly badger]

I then write this is equal to 1- 1/2^k then

[sly badger]

.

Would that be how my complete proof should look like

[quaint gulch]

for n = 1, the sum = 1-1/2 = 1/2, is base case
then from your induction, for all n, the sum is 1-1/2^n

so for all n it's smaller than 1

[sly badger]

Umm I'm lost

[quaint gulch]

why?

[sly badger]

I asked if my proof should look like this you are talking about proving the 1-1/2^k is in fact true for all n

So are you making me prove two induction statements in one or ?

[quaint gulch]

you want to prove $S(n) = \sum_{i = 1}^n \frac{1}{2^i} < 1$ for all n

[sly badger]

Yes

[quaint gulch]

so you can prove S(n) = 1-1/2^n for all n

[sly badger]

For some reason the statement says 1/2^i

[quaint gulch]

it follows that S(n) < 1 for all n, since 1-1/2^n is smaller than 1

[boreal girder]

Melo

[sly badger]

Oh I see

I see

That still means it smaller than 1

[quaint gulch]

by induction $S(n) = 1-\frac{1}{2^n}$

[sly badger]

But don't you have to say where S(n) is positive or something like that

[boreal girder]

Melo

[quaint gulch]

so S(n) < 1

[sly badger]

Yup

I understand it now

[quaint gulch]

hmm the exercise doesn't ask if S(n) is positive, but since it's a sum of positive terms it obviously is

so you can say it if you want

[sly badger]

Alright now I just prove this which is a much simpler proposition

And I already did

So thank you for your help I can finally sleep

It's 1 am

[compact pewter]

@sly badger Has your question been resolved?

Channel closed

Closed by @sly badger

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.

[quartz yoke]

how is this E??

[quartz yoke]

they are all multiples of 3 so you would assume they are kept 100%

[kind viper]

what do you mean by "they are kept 100%"?

@quartz yoke

alternatively: what can you place next to a multiple of 3 in the octagon?

[quartz yoke]

another multiple of 3?

I was thinking like 3 - 12 and 1 - 9 and 5 - 6 and 2 - 8 wouldnt that work?

oh right 2-8 doesnt work

but you get the idea how are they not used? 3 and 6 and 9 and 12

[kind viper]

5 & 6 don't work

[kind viper]

well you got it right, the only thing that can go next to a multiple of 3 is another multiple of 3

so you would need to fill the entire octagon with those

but you don't have enough for that

so you cannot use them at all

[quartz yoke]

cant I use the 4 multiples of 3 then the other 4 I use the rest?

[compact pewter]

@quartz yoke Has your question been resolved?

[kind viper]

no

because a multiple of 3 would have to go next to some non-multiple of 3

and you cannot have that

[compact pewter]

Channel closed

Closed due to timeout

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.

[dire rivet]

Okay so I was solving a question and got the answer wrong but then realized I would have gotten it right if it was 2x + 15 and that made me question that when you do lets say 180 - x -10 do you distribute the ("-") so it becomes 180 - x + 10 or it's just 180 - whatever the angle is

[dire rivet]

This is how I did it

[tired bear]

distribute it

say I take an angle 60

and then do 180 - 60

[dire rivet]

Oh so it's 180 - (2x -15)

[tired bear]

now i rewrite 60 as 50 + 10

yes

[dire rivet]

Ohhh

[tired bear]

you're removing both 50 and 10 from 180

[quartz yoke]

hi

[tired bear]

not just 50

[dire rivet]

Oh yeah

That makes sense

[tired bear]

so you remove every term in the angle from 180

and so you distribute the -

[dire rivet]

Thanks so much 🙏

Have a good day 😃

[quartz yoke]

is this not just no solution or am i tweaking

[tired bear]

ill have a good night instead 😄

you can close using .close

[quartz yoke]

damn.

i might just be tweaking.

[dire rivet]

Good night then 😃

[dire rivet]

?

[quartz yoke]

i thunk this is no solution for some reason

but i know thats wrong

[dire rivet]

What's no solution

[quartz yoke]

x

[dire rivet]

That's a part of the question not the question itself

[quartz yoke]

nvm

[dire rivet]

I did a very dumb mistake in that part

Yeah thanks for trying to help

[quartz yoke]

Aight bye

[dire rivet]

bye man have a good day

.close

[compact pewter]

Channel closed

Closed by @dire rivet

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.

[shadow igloo]

$z^3=-i$

[boreal girder]

Task Bot

Task Bot

Task Bot

Task Bot

[shadow igloo]

Correct?

[compact pewter]

@shadow igloo Has your question been resolved?

[compact pewter]

@shadow igloo Has your question been resolved?

[shadow igloo]

<@&286206848099549185>

[pure horizon]

no

[shadow igloo]

😭

[pure horizon]

shw ur work

[pure horizon]

Only this is right

[shadow igloo]

,rotate

[pure horizon]

Can u explain what you've done

[shadow igloo]

No

[pure horizon]

fair enough

wanna know my solution tho

[shadow igloo]

oh well I will fix it alone

[pure horizon]

damn

[shadow igloo]

for trolling

.close

[compact pewter]

Channel closed

Closed by @shadow igloo

Use .reopen if this was a mistake.

[shadow igloo]

Before I forget

<@&268886789983436800>

[snow finch]

what?

[shadow igloo]

Can you read the messages above? PLS

[thorny urchin]

It doesn't appear there was any trolling here.
It looked like they really wanted to help. (Though we don't support giving away solutions.)

[shadow igloo]

And where would this help be?

Only questions

[thorny urchin]

It is common practice to ask for work and reasoning.

If something like reasoning isn't provided as requested, there is less room for response.
Not sending messages isn't considered trolling, compared to intentionally spouting nonsense.

[shadow igloo]

Okay the fault is always mine

Now I go out of the discord that perhaps it is better

Bye

[pure horizon]

if u just wanted the solutions i could've provided u that

[thorny urchin]

Noone is saying its your fault here. If you have concerns, you are free to ping mods.

[pure horizon]

stop playing the victim card boi

all i said was that ur solutions are wrong

[thorny urchin]

But please don't freely give away worked solutions.

[pure horizon]

idc

[pure horizon]

when did i?

[thorny urchin]

It sounded like you were planning to.

[pure horizon]

i could've tell him the reasoning

leave this bs

[charred finch]

I remember that guy being in here since a long time, why did he get such a sudden mood swing lol

I'm sure he knows how it's done here

[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.

[fair raft]

Let $X$ be a set, $\mathcal{B}(X, \mathbb{R})$ the set of bounded functions from $X$ into $\mathbb{R}$ and $\forall f \in \mathcal{B}(X, \mathbb{R}), \ ||f||=sup_{x \in X} |f(x)|$.
Given this norm, show that $\mathcal{B}(X, \mathbb{R})$ is a Banach space.

[boreal girder]

tm

[fair raft]

so i have to show that for all cauchy sequences in B(X,R), it converges

but how i do that 😭

idk how to start

[pine osprey]

Mama les e.v normés

[fair raft]

😥

c’est harr

[pine osprey]

Soit fn une suite de cauchy de B(X,R)

Construit sa limite éventuelle

Vérifie que la limite est dans l'ensemble de départ et que la suite converge la bas

[fair raft]

mais si c’est de cauchy elle est bornée nan?

[pine osprey]

Bah tu as que pour a,b entiers |f_a(x) - f_b(x)| <= ||f_a - f_b|| et donc que c'est cauchy dans R

[fair raft]

😧

rien compris

😭

[pine osprey]

Et ducoup (f_n(x))_n converge

[fair raft]

ahhh mais nan t’as mit la norme mb

j’avais pas vu mdr

[fair raft]

oui

[pine osprey]

Oui

Mdr

[fair raft]

et faut montrer que c’est dans B truc ducoup

la limite

[fair raft]

ça doit pas être difficile à montrer que c’est borné

[pine osprey]

Tu l'as déjà fais

[pine osprey]

.

[fair raft]

ah oupsie

[fair raft]

mais ça ça vient d’où mdr

[pine osprey]

|f_n(x)| <= M

[fair raft]

c’est psk f(x) est dans R ?

fin f_a(x) quoi

[pine osprey]

Oui ça utilise le fait que R est complet

[fair raft]

rah ouais on peut pas faire de manière intrinsèque

[pine osprey]

Et ducoup avec f_n cauchy tu as ce truc

[fair raft]

ui

[pine osprey]

Mais ducoup avec ça

Tu passe à la limite

[fair raft]

la norme est continue donc on peut c’est ça ?

(en dehors de l’origine)

jcrois

jdis que de la merde je confond avec différentiable

<@&268886789983436800>

[pine osprey]

Non je voulais dire la valeur absolue mais en vrai tu as un truc comme ça : |f_n(x)| <= ||f_n|| <= M

Il veut quoi lahuiss

[fair raft]

mdrr gros scam sa mère

[fair raft]

ui mais où on passe à la limite la 😭😭😭

dans la valeur absolue ?

[pine osprey]

Partout en vrai non?

Puisque norm c'est sup de val abs

[fair raft]

bah c’est grâce a ça nan?

[pine osprey]

Oui

[fair raft]

oki

jvais essayer de mettre ca au propre mdrr

merci bg

[pine osprey]

Après la convergence tu fais avec epsilon

[fair raft]

le passage à la limite ?

[pine osprey]

Non après

Montrer que fn converge vers l

[fair raft]

ahhh

ui

donc att

faut montrer que ça converge

et que ça limite c’est dans B

les étapes pour montrer que c’est complet

[pine osprey]

Que ça converge dans B ouais

[fair raft]

oki

la prochaine fois j’éviterai de prendre un exercice du gourdon quand jcommence les espaces complets ..

[pine osprey]

Tu peux faire un truc avec les fonctions lipchitz bidules aussi, ça forge bien ça

Lipschitzienne

[azure vault]

alors alors ça avance cet exo ?

[pine osprey]

L'inspecteur est là tm vite, finit le boulot

[fair raft]

purée j’étais parti manger 😭😭

[fair raft]

le truc de la distance 1-lipschitzienne ?

ez ça

la c’est bon

[boreal girder]

tm

[fair raft]

est-ce que rafilou approuve la preuve ?

[azure vault]

"(f_n) sont aussi de cauchy dans R"

il manque un truc?

[fair raft]

c’est a dire ?

[azure vault]

(f_n) c'est une suite de fonctions

[fair raft]

(f_n(x)) ?

[azure vault]

voilà

[fair raft]

pour x dans X

[azure vault]

oui c'est mieux

[fair raft]

erreur d’inattention 😔

[azure vault]

il manque le quantifieur pour x ici

[fair raft]

oui bon encore erreur d’inattention😭

ça fait beaucoup

[azure vault]

oui, et c'est là que ça va devenir chaud je pense

[fair raft]

mais j’le met après pour tout n,m ?

fin je dirais non

psk sinon c’est uniforme

[azure vault]

je te laisse mettre ça au clair

parce qu'en suite y a un problème bcp plus gros

[fair raft]

jcrois voir de quoi tu parles mdr

le passage au sup?

[azure vault]

bah même avant

comment tu justifies la majoration

[fair raft]

$|f_n(x)-f(x)|=|f_n(x)-f_m(x)+f_m(x)-f(x)|$

[boreal girder]

tm

[fair raft]

ou alors je suis fou

[azure vault]

c'est pas ça le problème

inégalité triangulaire oui

après pourquoi 2epsilon?

[fair raft]

o le con

[azure vault]

et je viens de capter mais

[fair raft]

je me disais |f_m(x)-f(x)| c’était borné dcp ça passe

[azure vault]

non ça c'est bon

[fair raft]

eh oh je fais pas 4 erreurs par ligne quand même 😔

[azure vault]

oui ça ça reste le problème

[fair raft]

on passe a la limite alors

[azure vault]

donc en fait pour majorer |f(x) - f_n(x)|

[fair raft]

pour f_n ou f_m

[azure vault]

faut majorer |f(x)-f_m(x)|

donc inutile

[fair raft]

ui ui

[azure vault]

autant commencer de:
$$\forall x\in X, ,\forall m,n\geq N, ; |f_n(x)-f_m(x)| < \varepsilon$$

[boreal girder]

rafilou is not not born in 2003

[azure vault]

bon de ça effectivement

[pine osprey]

n --> + inf ?

[fair raft]

on fait tendre m vers l’infini ?

ou n ouais

[azure vault]

oui l'un ou l'autre

et l'inégalité devient large

mais osef

[fair raft]

ah rip

[pine osprey]

Puisque c'est vrai pour tout x dans X

[azure vault]

on en déduit alors que f_n - f est bornée

[fair raft]

ui

[azure vault]

(bon ça en vrai c'était un peu couru d'avance)

et ||f_n - f|| <= epsilon pour n >= N

bon bah on a gagné

[pine osprey]

1-0

[fair raft]

mais on a pas construit f pour que ça soit la limite de f_n ?

[azure vault]

oui mais f c'est la limite simple de (f_n)

[pine osprey]

C'est juste un nom i bet

[azure vault]

la limite point par point

[fair raft]

a faut aussi montrer la convergence en norme

[azure vault]

bah oui

eh

[fair raft]

rah ouais relou les espaces complets en vrai

[azure vault]

si t'as convergence simple et convergence norme infini, alors oui les deux limites sont pareilles

d'ailleurs t'as même convergence norme infinie -> convergence simple

[fair raft]

ui c’est sur

[azure vault]

mais ayant juste convergence simple

aucune raison qu'on ait convergence en norme infinie

exemple: f_n(x) = delta(x,n)

[fair raft]

uh c’est quoi ça delta mdr

[azure vault]

delta étant symbole de kronecker bien sûr

[fair raft]

aaa

u

i

[fair raft]

c’est 0 si x=n et 1 si x≠n ?

[azure vault]

c l'inverse

[fair raft]

oki

ah oupsi

[azure vault]

donc bref f_n converge point par point vers 0 (f_n(x) -> 0 pour tout x)

et pourtant

la norme de la diff c'est toujours 1

[fair raft]

avec la norme sup

[azure vault]

oui norme sup/infinie

[fair raft]

hmm je vois

bon jvais jouer a rocket league les maths c’est fini ce soir mdr

merci 🙏🏼

.close

[compact pewter]

Channel closed

Closed by @fair raft

Use .reopen if this was a mistake.

[azure vault]

🚗 ⚽ 🚙

bonne soirée xd

[pine osprey]

Mesure la norme pour faire des aerials

[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.

[cosmic geode]

i know how to get 1, and infer that because its an absolute value function, the reverse sign of that -1 must also be a critical number. I dont get how x=0 came about though

[wise jolt]

x=0 the derivative doesnt exist so it counts as a critical point

[cosmic geode]

is there a way to recognize that? or should i just always plug in x=0 to check

[wise jolt]

in this case you can recognize it by the fact that its absolute value

[cosmic geode]

okay cool

[wise jolt]

if its some random piecewise function then you could check first if its continuous at that point by seeing if the limits are equal and also check that the derivatives are equal

[cosmic geode]

thank you

[wise jolt]

np

[cosmic geode]

.close

[compact pewter]

Channel closed

Closed by @cosmic geode

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.

[shadow igloo]

$D\cos(x)^\frac{1}{x}$

[boreal girder]

Chicken Jockey🐔

[shadow igloo]

Pls help me !

[lean berry]

[shadow igloo]

?

Help me

[lean berry]

What's ur question?

[shadow igloo]

Pls

I have to do the thing above

[lean berry]

What is D?

[shadow igloo]

D : = derivatives

Thanks

[lean berry]

Owh no

Nooo

U are university guy

[shadow igloo]

No

[lean berry]

My level is lowest than urs

It's wrong answer

Add ln

[shadow igloo]

How do I solve the question?

[lean berry]

f(x) = cos^1/x
ln(f) = 1/x ln(cos(x))

[shadow igloo]

I can translate into latex

$f(x) = \cos^\frac{1}{x}
\ln(f) = \frac{1}{x} \ln(\cos(x))$

[boreal girder]

Chicken Jockey🐔

[lean berry]

$\frac{f'}{f} = \frac{-1}{x^2} \cdot ln(\cos(x)) - \frac{\tan(x)}{x}$

[boreal girder]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[lean berry]

Multiply both sides by cos(x)^{1/x}

[shadow igloo]

$\frac{f'}{f} = \frac{-1}{x^2} \cdot \ln(\cos(x)) - \frac{\tan(x)}{x}$

[boreal girder]

Chicken Jockey🐔

[shadow igloo]

$\cos(x)^\frac{1}{x}\frac{f'}{f} = [\frac{-1}{x^2} \cdot \ln(\cos(x)) - \frac{\tan(x)}{x}]\cdot \cos(x)^\frac{1}{x}$

[boreal girder]

Chicken Jockey🐔

[shadow igloo]

Then?

[lean berry]

bruh

f = cos(x)^1/x

[shadow igloo]

$f' = [\frac{-1}{x^2} \cdot \ln(\cos(x)) - \frac{\tan(x)}{x}]\cdot \cos(x)^\frac{1}{x}$

[boreal girder]

Chicken Jockey🐔

[lean berry]

Great, u have reached ur goal

[shadow igloo]

Thanks

[lean berry]

!done

[compact pewter]

If you are done with this channel, please mark your problem as solved by typing .close

[shadow igloo]

.close

[compact pewter]

Channel closed

Closed by @shadow igloo

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.

[reef bronze]

i'm looking for primes p such that n^4+1 is divisible by p^2 for some integer n
that would mean n^8 is congruent to 1 mod p^2, so the order of n is 8
so 8 divides phi(p^2)=p(p-1)
so the smallest candidate for p is 17
does that make sense?

[reef bronze]

oh that only works for n coprime to p^2 but for the other n we have n^4 congruent to 0 mod p^2 so those don't work

[quartz yoke]

why don't you take order of n as 4 instead of squaring it

[reef bronze]

the order of n can't be 4 because n^4 is congruent to -1

[quartz yoke]

oh my bad lmao

[reef bronze]

no worries

[compact pewter]

@reef bronze Has your question been resolved?

[reef bronze]

.close

[compact pewter]

Channel closed

Closed by @reef bronze

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.

[quartz yoke]

Hey all, this is how I started things off but am not sure how to compute the determinant on the last line (since it's a block matrix), thanks for your help

[static timber]

If $A$ is a block matrix of the form $A = \begin{bmatrix}B~0 \ 0~I\end{bmatrix}$, where $I$ is an identity matrix, then $\operatorname{det}(A) = \operatorname{det}(B)$.

[boreal girder]

ucheo

[quartz yoke]

I thought of trying to use the following property

Let me know if it's suitable here

Theorem: The determinant of an upper triangular or lower triangular matrix is the product of it's diagonal entries

Because

[static timber]

Yes, in this case that works. In particular, the determinant of any identity matrix is 1 Since $\begin{bmatrix}I_n~0 \ 0~I_m\end{bmatrix}$ is itself an identity matrix, its determinant must be 1.

[quartz yoke]

Isn't the matrix simply diagonal actually?

[boreal girder]

ucheo

[quartz yoke]

This is what I have

I think it SHOULD work?

[static timber]

Yeah, looks correct

[compact pewter]

@quartz yoke Has your question been resolved?

[quartz yoke]

Thank you!

.close

[compact pewter]

Channel closed

Closed by @tiny gull

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.

[limber spade]

how do i solve this?

[charred finch]

Dropping perpendiculars from the point into the x and y axes might help

[inland nacelle]

Do you know that the equation of a circle is (sin theta, cos theta)

Nah @charred finch method is better

[limber spade]

idk what that means

im kinda lost

am i supposed to find the angle first

[charred finch]

Draw this

Where they r perpendicular to the x and y axes

[charred finch]

Let x be this angle

We get tan x = 0.8/0.6 = 4/3 => x ≈ 53°

[real magnet]

The y coordinate gives the value of sin(theta) and the x coordinate the value of cos(theta).

[charred finch]

Hence theta = 360° - 53° = 307°

[real magnet]

What?

[charred finch]

Mb

[limber spade]

i have no clue what any of that means

we did not go over any of this in class

what is sin and cos supposed to be

[charred finch]

U mean u want to know what are the sin and cos functions or the value of them in this case?

[limber spade]

do i put sin and cos (307) into the answer boxes?

[charred finch]

Yes

[limber spade]

i did and it said it was wrong?

[charred finch]

Honestly, the last step is not necessary

U need to simplify it

We get tan x = 4/3 right

Do u know how this came?

[limber spade]

0.8/0.6

[charred finch]

Yes

And we know $\theta+x=360^{\circ}$

[boreal girder]

Suika

[charred finch]

Hence $x=360^{\circ}-\theta$

[boreal girder]

Suika

[charred finch]

$\tan{x}=\tan{(360^{\circ}-\theta)}=-\tan{\theta}$

[boreal girder]

Suika

[charred finch]

Hence $-\tan{\theta}=\frac{4}{3}$

[boreal girder]

Suika

[charred finch]

Tan theta is -4/3

From that, we get sin theta is -4/5 and cos theta is 3/5

Honestly, I'm shit

Ask someone else to explain and forget whatever I said

[limber spade]

i dont understand what the question is asking for at all

like what does it want

[charred finch]

No, cos is 0.6 and sin is -0.8

[limber spade]

thats so dumb

i know how to do this with a protractor how i am i supposed to do it without

[pearl sierra]

Drawing the unit circle would help

And then using some neat symmetry properties

[pearl sierra]

On the unit circle like here we the x values are cos(x) and the y values are sin(x)

[compact pewter]

@limber spade Has your question been resolved?

Channel closed

Closed by @limber spade

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.

[quick jewel]

Proof

[quick jewel]

Derive the formula for $\sum_{x=1}^{n} x^4$ using the identity $x^5 - (x-1)^5$.

• Identity: $x^5 - (x-1)^5$

Helpful information

• Summation formulas:
• $\sum_{x=1}^{n} 1 = n$
• $\sum_{x=1}^{n} x = \frac{n(n+1)}{2}$
• $\sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}$
• $\sum_{x=1}^{n} x^3 = \frac{n^2(n+1)^2}{4}$ [1]

1. Step 1 Expand $(x-1)^5$
   • Using the binomial theorem:
   • $(x-1)^5 = x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1$

2. Step 2 Simplify $x^5 - (x-1)^5$
   • $x^5 - (x-1)^5 = x^5 - (x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1)$
   • $x^5 - (x-1)^5 = 5x^4 - 10x^3 + 10x^2 - 5x + 1$

3. Step 3 Sum both sides from $x=1$ to $n$
   • $\sum_{x=1}^{n} (x^5 - (x-1)^5) = \sum_{x=1}^{n} (5x^4 - 10x^3 + 10x^2 - 5x + 1)$

4. Step 4 Evaluate the left side of the equation
   • $\sum_{x=1}^{n} (x^5 - (x-1)^5) = (1^5 - 0^5) + (2^5 - 1^5) + \dots + (n^5 - (n-1)^5)$
   • This is a telescoping sum, so all terms cancel except the last one:
   • $\sum_{x=1}^{n} (x^5 - (x-1)^5) = n^5$ [1]

5. Step 5 Evaluate the right side of the equation

• $\sum_{x=1}^{n} (5x^4 - 10x^3 + 10x^2 - 5x + 1) = 5\sum_{x=1}^{n} x^4 - 10\sum_{x=1}^{n} x^3 + 10\sum_{x=1}^{n} x^2 - 5\sum_{x=1}^{n} x + \sum_{x=1}^{n} 1$ [1]
• Substitute the known summation formulas:
• $5\sum_{x=1}^{n} x^4 - 10\frac{n^2(n+1)^2}{4} + 10\frac{n(n+1)(2n+1)}{6} - 5\frac{n(n+1)}{2} + n$ [1]
• $5\sum_{x=1}^{n} x^4 - \frac{5}{2}n^2(n+1)^2 + \frac{5}{3}n(n+1)(2n+1) - \frac{5}{2}n(n+1) + n$ [1]

6. Step 6 Set the left and right sides equal and solve for $\sum_{x=1}^{n} x^4$
   • $n^5 = 5\sum_{x=1}^{n} x^4 - \frac{5}{2}n^2(n+1)^2 + \frac{5}{3}n(n+1)(2n+1) - \frac{5}{2}n(n+1) + n$ [1]
   • $5\sum_{x=1}^{n} x^4 = n^5 + \frac{5}{2}n^2(n+1)^2 - \frac{5}{3}n(n+1)(2n+1) + \frac{5}{2}n(n+1) - n$ [1]
   • $5\sum_{x=1}^{n} x^4 = \frac{30n^5 + 75n^4 + 110n^3 + 60n^2 + 5n}{30}$
   • $\sum_{x=1}^{n} x^4 = \frac{6n^5 + 15n^4 + 10n^3 - n}{30}$
   • $\sum_{x=1}^{n} x^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

[boreal girder]

Salik

Salik

[quick jewel]

Someone pls explain how we reached from second last to last step in this photo

• $\sum_{x=1}^{n} x^4 = \frac{6n^5 + 15n^4 + 10n^3 - n}{30}$
• $\sum_{x=1}^{n} x^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

[boreal girder]

Salik

[kind viper]

where did you copy and paste this from

anyway this looks like they did some factorization magic

[quick jewel]

From google

[kind viper]

what like directly from google

or did you follow a link in a search result

[quick jewel]

Ya but I don't know what happened afterwards

[kind viper]

wdym afterwards

this is the last step by the looks of it

there's no afterwards

[quick jewel]

First I solved in notebook then I reached the second last step
But couldn't know how to factorise

Then I searched on Google and it showed no factorization step
It just gave direct answer after the second last step

[kind viper]

well ok like

[quick jewel]

Idk what happened in this
Like how do I factorise it?

[kind viper]

so you understand all of the other stuff here and your issue is specifically about how to factorize 6n^5 + 15n^4 + 10n^3 - n

[quick jewel]

Yes and then finally get the last step after factorization

[kind viper]

n can be factored out right away as you can hopefully see

[quick jewel]

Hmm

[kind viper]

after that, i would try the rational root theorem to get some nice roots and thus nice corresponding factors

[quick jewel]

What is this rational root theorem 😭

[kind viper]

https://www.cuemath.com/algebra/rational-root-theorem/
https://en.wikipedia.org/wiki/Rational_root_theorem

[quick jewel]

I am from India
Never heard of this

[kind viper]

you can also google it :PPPPP

[quick jewel]

Hmm

[kind viper]

the rational root theorem says that if a polynomial has integer coefficients then its rational roots will only take a specific form

which can help you either find them all in a not too big search space or be assured that if you didn't find any then you couldn't have done so anyway

[quick jewel]

Oh 🙃

Will try

Thank You Very Much for the help

@kind viper

[compact pewter]

@quick jewel 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.

[true coral]

Can some1 help me with this???

[kind viper]

,rccw

[kind viper]

what are you asked to do w this

[trail cave]

i cant read that sry whats the denominator

[kind viper]

also that fraction bar is so short...

[true coral]

Wait

[kind viper]

$f(x)=\frac{4x+1}{x^2-kx+k}$

[boreal girder]

Ann

[true coral]

f(x) = 4x -1/ x^2 - kx + k

Yes!

[kind viper]

you need brackets in plain text

anyway

what are you asked to do with this

[true coral]

I was meant to do delta > 0

[gilded kayak]

delta of f(x)?

is this even quadratic