help-49

[novel herald]

x concatenated with ()?

[floral minnow]

which x?

[novel herald]

doesn't matter

[floral minnow]

x E S?

[novel herald]

if x is a string in S

what is (x) supposed to be

[floral minnow]

any item from the set S?

i think

[novel herald]

i mean i don't understand how they're defining the opertions of S but

[floral minnow]

oh

im not sure either tbh sorry

[novel herald]

i would assume that x and (x) are distinct, and nowhere does it say that if x is in S then (x) is as well

so choice b, (c-d) shouldn't be in S

choice c, (a) isn't in S

and choice d, (c) isn't in S

[floral minnow]

and thats because the subtraction rule doesnt have paratheses?

[novel herald]

yes

[floral minnow]

so its the parentheses that are messing me up

ok

[novel herald]

c,d in S by rule 2, c-d is in S but (c-d) isn't

weird problem though

that's my guess atleast

[floral minnow]

so for this example

it wouldnt be a) because its in parenthes?

but like a-b-c would be?

[novel herald]

a - b - c would be based off of the rules

a and b are in S so by rule 2, a - b is in S

then by rule 2 again a - b - c is in S

[floral minnow]

so a-b-c is NOT the same as (a-b-c)?

[novel herald]

i am assuming they are different

[floral minnow]

thats so weird

ok

thanks

[novel herald]

but again without clarification idk

[floral minnow]

tysm

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[midnight plank]

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[floral minnow]

nice pfp btw

[midnight plank]

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[wary thorn]

⚔️ ⚔️ ⚔️

[carmine void]

What do you think?

[bleak pier]

I took some random polynomials

[tidal token]

weird question ngl

[olive matrix]

it's a good question

[tidal token]

yeh i think it's c but how do u prove it

ik the proof isnt necessary for this q

but im intrigued ngl

[olive matrix]

it's a fairly straightforward proof; i think you can do it

[tidal token]

for any polynomial, f ? of any degree? nah i cant prove it

[bleak pier]

Where should I start from?

[olive matrix]

well first provide a counterexample for (a)

[gleaming latch]

proof by contradiction? idk

[olive matrix]

yes proof by contradiction or contrapositive is a good way to go here

[gleaming latch]

oh yeah i guess if you provide counterexamples for the other 3 works

[olive matrix]

i'd say contrapositive

[tidal token]

doesn't that just prove that the other 3 arent true?

[bleak pier]

For A (x-3)^3

[gleaming latch]

then the remaining one has to be true

[tidal token]

yes but how do i prove it outside the context of the question

[olive matrix]

use contrpositive

[gleaming latch]

prove that if a polynomial has a positive root then the polynomial must have a negative coefficient

[bleak pier]

.

[gleaming latch]

wdym

[olive matrix]

yeah good that's a counterexample for a

[gleaming latch]

oh ok

[olive matrix]

so now you have counterexamples for a, b, and d

so for this question you're done

[bleak pier]

How will we discard option b?

[tidal token]

contrapositive of this statement is: "If all cooefficients of the polynomial f(x) are negative, then f(x) does not have a negative root ."

this is just as hard to prove as the original statement

[gleaming latch]

(x+1)^3?

[olive matrix]

looks like you already did?

or just x+1 lol

[gleaming latch]

[bleak pier]

Statement is if coffiecients are positive then equation will not have a reapeated root

(x-3)^3
Repeated roots and all coffiecients are not positive

[olive matrix]

yes good

wait... no, that doesn't help you much

you need to find something that has all positive coefs and has a repeated root

[bleak pier]

Then how is it good example?

[olive matrix]

it's a counterexample

you're trying to disprove the statement

[bleak pier]

I do not think my example was counter

[olive matrix]

you're right it wasn't

you need to find a counterexample

[bleak pier]

Let me think

[olive matrix]

i misread earlier when i said it was good

[bleak pier]

(x+1)^3

Here roots are reapeated

And all coffiecients are positive

[olive matrix]

yes good

[bleak pier]

For B
(X+4)^2
Roots negative and coffiecients are positive

It will be option C

Because the negatuve term will make ome term negative so all the coffiecients can not be positive

Thank you all

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[thick shuttle]

What's the limit definition of the 2nd derivative if you already have the 1st?

[visual tiger]

the second derivative is the derivative of the derivative
f"(a) = (f')'(a)

So if you recall the definition of the first derivative

[thick shuttle]

yep

[visual tiger]

Just... swap the f with f'

[thick shuttle]

oh really?

thanks!

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[graceful rain]

how do i approach this kind of question?

[graceful rain]

p-series test says that if p > 1 then the series converges, if less than or equal to it diverges

is there any way to solve for p algebraically?

[midnight plank]

@graceful rain Has your question been resolved?

[graceful rain]

<@&286206848099549185>

[midnight plank]

@graceful rain Has your question been resolved?

[lapis osprey]

Have you tried evaluating the integral?

[graceful rain]

I haven't

[midnight plank]

@graceful rain Has your question been resolved?

[rotund granite]

Did you try resulting the c factor?

[loud vortex]

P sure you dont need to evaluate the integral

What you had with the p series test should suffice

p/3 >1 for convergence

[valid shuttle]

@graceful rain do you still need help?

[graceful rain]

yes please

[valid shuttle]

What does the p value need to be in order for the series to diverge

Converge*

[graceful rain]

greater than 1 sorry

[midnight plank]

@graceful rain Has your question been resolved?

[normal moat]

So let p’=p/3. p’ needs to be >1 for convergence. So p/3>1 solve for p

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[midnight plank]

Please don't occupy multiple help channels.

[midnight plank]

@worn relic Has your question been resolved?

[midnight plank]

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[steady trail]

A straight wire AB of length L and negligible cross section, arranged along an x ​​axis as shown in the figure, is
uniformly charged with linear charge density $\lambda$. Calculate the electrostatic field generated by the wire at a point C
placed on an axis perpendicular to the wire, at a distance h from the wire itself and at a point D on the x axis located at a distance L
from end B of the wire.

[grand pond]

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

[steady trail]

pls help

[midnight plank]

@steady trail Has your question been resolved?

[steady trail]

<@&286206848099549185>

[hard jolt]

Yea

[steady trail]

👋

<@&286206848099549185>

[pulsar sun]

It's physics

I don't think there will be people that will answer you

[steady trail]

ok, sorry

[hard jolt]

Yea ..

[steady trail]

👍

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[shrewd finch]

can someone help me out with this

[rose cobalt]

Where do you need help

[midnight plank]

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[shrewd finch]

how do i solve it

[rose cobalt]

Do you know the trapezium rule?

[rose cobalt]

.

[shrewd finch]

yeah

why do we need the trapezium rule for this

is there no other way

[midnight plank]

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[steep quest]

how does this go to this

[steep quest]

was this rationalization or?

[midnight plank]

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[gleaming spear]

Simplify
[\cos \frac{2 \pi}{13} + \cos \frac{6 \pi}{13} + \cos \frac{8 \pi}{13}.]

[grand pond]

Dork9399

[gleaming spear]

@hearty rune what do you think

[hearty rune]

i think of addition formulae

[gleaming spear]

the tan addition?

or cos addition

[charred plinth]

cos thing right

cos(theta) + cos(60 - theta) + cos (60 + theta)

but im not sure on how you get that to this format

[gleaming spear]

yea how will that work for this

@hearty rune any ideas

can I use roots of unity somehow

[hearty rune]

cos(8pi/13)=cos(2pi/13)cos(6pi/13)-sin(2pi/13)sin(6pi/13)
cos(6pi/13)=cos(2pi/13)cos(4pi/13)-sin(2pi/13)sin(4pi/13)
you can probably just keep breaking it down

[hearty rune]

possibly

[gleaming spear]

I can also just use triple and quad angle

but how would that help

it simplifies to this

8cos^4(theta) + 4cos^3(theta) - 8cos^2(theta) - 2cos(theta) + 1

[manic plover]

you could add the first and the last term using c + c formula

you would get cos 6 pi/13 common however i'm not sure how to proceed further

[manic plover]

cos a + cos b = 2 cos [(a+b)/2] * cos [(a-b)/2]

[midnight plank]

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[gleaming spear]

.close

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[midnight plank]

@bleak pier Has your question been resolved?

[bleak pier]

<@&286206848099549185>

[whole escarp]

can you type it

[bleak pier]

what should I type

[whole escarp]

this

[remote sand]

According to option A when you find the second derivative and put the double root is it you'll get 0

[whole escarp]

im gonna leave this to gaza on second though

[bleak pier]

i checked all the options and got all wrong

[remote sand]

Try retracing them you might have done something wrong

[midnight plank]

@bleak pier Has your question been resolved?

[grand pond]

Dyssrupt

[ebon sable]

now try differentiating that using product and chain rule and find f'(x)

[bleak pier]

2( x-alpha)(x-beta)+(x-alpha)^2

(x-alpha)(2x-2beta+x-alpha)
x-alpha(x-2 betw-alpha)

[ebon sable]

try x = alpha there :)

[bleak pier]

It will be 0

[ebon sable]

so alpha is also a root of f'(x) = 0

so you have a general result now

[bleak pier]

(x-2)^2(x-3)

You are right

Thanks

[ebon sable]

if a polynomial has a repeated root, it will also be the root of its derivative

[midnight plank]

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[midnight plank]

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[drowsy thunder]

Is my approach for C correct

[hearty rune]

uh, i dont believe so

im curious where these -5s came from too

ah .5s

lets see then

-(0 to 0.5) of 2g

about g you know

-1 to .5 is -3/4
2 to .5 is 8
0 to 2 is -4

[drowsy thunder]

what about the 2 in 2g(x)

[hearty rune]

what about it?

[drowsy thunder]

oh wait did u divide my -2?

to get 3/4

[hearty rune]

yeah

[drowsy thunder]

i see

thx

.close

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[last slate]

How do i find a closed form formula for this?

[rose trout]

Can't you just expand it straight away?

[last slate]

?

[rose trout]

-9 + 27, ...

[last slate]

has to be a formula

[rose trout]

Why so

[radiant trout]

geometric series

[rose trout]

Yeah the geometric series formula would work, just sounds a bit overkill for 6 terms.

[last slate]

that works when its alternating?

[rose trout]

Just put r = -3

And adjust the first term since j starts at 2

[last slate]

ty

so something like this

thats crazy

[glad tree]

hello

[last slate]

Hey, how are you?

[glad tree]

am good how are you

[last slate]

I'm horrible. Born to die, forced to do math

[glad tree]

da,n

*damn

how many ticcles does it take to make a octous laugh

[midnight plank]

@last slate Has your question been resolved?

[last slate]

at least 8

[midnight plank]

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[spark pewter]

Use proof by induction to prove that the sum of the first n terms of an arithmetic sequence is given by the formula below.
s subscript n = n/2 x [2a+(n-1)d]

What does this question mean?

[olive shoal]

do you know what proof by induction is?

[spark pewter]

Yes, you prove the base case, than assume for K is true. than do k+1 and subsitute K into it

[olive shoal]

yeah so do that

[spark pewter]

But I am not understanding the question

'Prove the sum of the first n terms of an arithmetic sequence'

Am I proving that any sequence will fit into that formula

[exotic musk]

yes

the base case is straightforward

have you completed showing the base case

[spark pewter]

1,2,3,4,5?

[exotic musk]

no

n=1

[spark pewter]

Which = a

[exotic musk]

yep

now assume true for n = k

[spark pewter]

Than do K?

[exotic musk]

yep

[spark pewter]

So S(k) = k/2(2a+(k-1)xd)

[exotic musk]

yep

[spark pewter]

Then K+1

[exotic musk]

now show true for n = k +1

yep its just the normal proof by induction

[spark pewter]

What I am struggling with is how to subsitute the K into the K+1

I don't need to simplify at all?

[exotic musk]

so what would S(k+1) be in terms of S(k)

[spark pewter]

And shouldn't this have an answer?

Like an =

[exotic musk]

no, that's what your assuming true

[spark pewter]

Ok so once I have Replced the K in K+1 with k/2(2a+(k-1)xd), I do what?

I am definitly missing something here

Like knowledge wise

[exotic musk]

you dont replace anything

[spark pewter]

Ojhhhhhh

Wait

I forgot it's a sequence

[exotic musk]

using knowledge of an arithmetic sequence and S(k) show S(k+1) in a similar form to S(k)

[spark pewter]

I'm still confused

[exotic musk]

how does S(k+1) relate to S(k)

just generally

[spark pewter]

It would but Idk in what way

Coz it's for any sequence so

[exotic musk]

yh just generally, how would you get S(k+1) from S(k)

think about it

[spark pewter]

By adding 1

[exotic musk]

no

[spark pewter]

Subbing one sorry

[exotic musk]

by adding the k+1 th term

[spark pewter]

?

To get k+1 to K you'd subtracting 1

I am so confused rn

[exotic musk]

thats just k and k+1

[spark pewter]

Yeah

[exotic musk]

read this again

what is S?

[spark pewter]

The term

[exotic musk]

no

have you read the question

[spark pewter]

Yes, and I don't understand it

[exotic musk]

so you don't know what S is?

hint (it's in the question you sent)

[spark pewter]

The sum

[exotic musk]

so try this again

[spark pewter]

Wait so the question is saying, for a sequence, for example, {2,4,6,8,10...}, the sum of the first n terms, for example the first three will equal 3/2{2a+(2-1)d}?

[spark pewter]

So the sum of K+1

[exotic musk]

are you sure this is right: 3/2{2a+(2-1)d}?

[spark pewter]

No 😭

[exotic musk]

trying subbing n again

[spark pewter]

wdym

[exotic musk]

How did you get this 3/2{2a+(2-1)d}?

[spark pewter]

Using the sequence

Of the first 3 terms

which was, 2,4,8

[exotic musk]

so you didnt use the formula given in the question

[spark pewter]

I did?

Or I think I did, now Im questioning everything

[exotic musk]

so you put n=3 in this formula?

[spark pewter]

Yes

[exotic musk]

and got 3/2{2a+(2-1)d}

[spark pewter]

Yes

[exotic musk]

try again

[spark pewter]

S(3) = 3/2{2a+(3-1)d}?

So this isn't right

Wait Isee

I did 2 instead of 3

[exotic musk]

yes

[spark pewter]

What's a and d?

[exotic musk]

so the first 3 terms will be equal to this

[exotic musk]

what do you think / remember?

[spark pewter]

?

Is this arithmetic formula

wait d is the common difference

[exotic musk]

revisit arithmetic sequences if unsure

[spark pewter]

I know d is the differnece, so two in this case. but idk what a is

The first term I would've thought

Except when I did 12 = 2/2(2a+(2-1)x2) a=3 and the first term was 2

[exotic musk]

it is

[spark pewter]

Wait I put in the wrong nubers again

Bro your probably so sick of me rn

So,
S(k+1) = (K+1)/2(2a+Kd)

[exotic musk]

yep that should be your end result, not what you start with

[spark pewter]

How does that prove it tho

[exotic musk]

if it works for n= 1

and we assume it true for n= k

then show it works for n= k +1

we can start for n = 1, which we know is 100% true

then we say n = 2 works, n =3 works so on and so forth

[spark pewter]

But in my head we haven't worked out that k+1 is actually equal to that. I just shoved in the k+1 into the formula

[exotic musk]

yh ur not supposed to do that in induction

[spark pewter]

Aren't you? We havent proved K+1 works

And we assumed K works

Only 1 works

[exotic musk]

revisit proof by induction

[exotic musk]

ur supposed to do this now

[spark pewter]

😭

[exotic musk]

anyways its been 40 mins i gtg. If you're struggling this much I suggest you much some yt vids on arithmetic sequences and proof by induction on sequences

[spark pewter]

I have 😭😭

[spark pewter]

Any thanks for your help

<@&286206848099549185> I think I am allowed to do this now. Is there anyone else who can help.
Question: Use proof by induction to prove that the sum of the first n terms of an arithmetic sequence is given by the formula below.
s subscript n = n/2 x [2a+(n-1)d]

So far,
S(1)= 1/2(2a+(1-1)xd) = a
S(k) = k/2(2a+(k-1)xd Assume this is true
S(k+1) = (k+1)/2(2a+(k+1-1)d)
=(k+1)/2(2a+kd)

Now I am a little stuck on how we use the S(k) we assumed is true to prove the S(k+1) works

[midnight plank]

@spark pewter Has your question been resolved?

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[spark pewter]

.reopen

[midnight plank]

✅

[spark pewter]

<@&286206848099549185>

[tribal temple]

Note that S(k + 1) = S(k) + (a + kd)

(Take the sum of the first k terms, then add the (k + 1)th term that you’re missing to get the sum of the first (k + 1) terms)

From where it should hopefully be clear as to where you use the induction hypothesis

[spark pewter]

Why adding a+kd

[deep quail]

that is the value of (K+1 )th term
a(k+1)th = a + (K + 1 - 1)d

[spark pewter]

So a+kd is equal to 1?

[deep quail]

no-

ok imagine this AP:

a,a+d, a+2d,...... a+kd

now the last term is (k+1)th term in this ap right?

[spark pewter]

Wdym by ap

[deep quail]

arithmetic progression- the sequence

[spark pewter]

Oh that's just a sequence

[deep quail]

yes yes...

tell me what do you understand by S(k)?

it is the sum of all?

[spark pewter]

The sum of the k terms equal that formula I was given

[deep quail]

yes

now if we add another term which is after the k term

that is (k+1) th term

[spark pewter]

Agreed

[deep quail]

we get S(k+1)

[spark pewter]

Yep

[deep quail]

which is the sum of the k+1 terms

so know

S(k+1) = S(k) + the (k+1)th term

[spark pewter]

Yes...

I think I see where this is going

[deep quail]

the (k+1)th term = a + (k+1-1)d

since for nth term = a + (n-1)d

[spark pewter]

Ohhhh I think I get it

When I get home and I'll do it and see if it's right

[deep quail]

Sure :D

[spark pewter]

Thanks soo much

[deep quail]

you're most welcome

[spark pewter]

Induction is such an interesting proof

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[tender cypress]

Hello

[grand pond]

nyes1375
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tender cypress]

How did they get the B(2x+1)+C??

or just the 2x+1

[slender walrus]

note that 2x+1 is the derivative of the denominator

B(2x+1) + C is still linear,
they're just expressing it like this to save some work later

as after this you'll be able to separate the part with the B(2x+1) into a log integral

[tender cypress]

Yepp, thank you very muchh

.close

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[olive matrix]

why have i never thought to do that

[slender walrus]

same

mindblown

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[dry oasis]

Hello! Pls help roughly calcualte lg7 (iirc with the error of 0.001) using Series

[dry oasis]

I tried going with the base of e, like ln7/ln10

But then realised that I can't use Taylor's series for ln(1+x), x here is from -1 to 1

So I can pull out a n8mber and use it using log property

but then I would have ln6 above and ln9 below

So it's stuck procedure I guess

[bold beacon]

You can do the taylor series at x = 6. I remember there being youtube video on calculating logs in SOME3 or SOME2

[dry oasis]

Huh. really? I think x is only -1 to 1

[bold beacon]

radius of convergence will depend on what what point is selected

[bold beacon]

you are selecting a particular point i.e. x=0 for the taylor's series of ln(1+x). But that is not necessary.

[dry oasis]

And this point must be between -1 and 1

[bold beacon]

Watch the youtube video then

[dry oasis]

If it's not between these points it doesn't work

If your |x| is more than 1 it won't converge

[bold beacon]

You don't seem to have understood what I have said. I will leave you to it

[dry oasis]

You basically told me to put 6

Yeah, hopefully gonna get right people😛

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[sturdy crystal]

is the probability of an event zero when the sample space iis infinite???

[sturdy crystal]

infinite possible outcomes i mean

[surreal moon]

Not necessarily, no

[sturdy crystal]

@surreal moon

[surreal moon]

Consider $\Omega=\left{E_n=\frac{1}{2^n} :n\in\bN\right}$

[sturdy crystal]

there is a fault in the 1/inifnity = 0 thing then

omega is the probability?

[grand pond]

SWR

[surreal moon]

Omega is the universe of all events

[sturdy crystal]

sample space you mean

[surreal moon]

There are an infinite number of events, each with nonzero probability, and all sum to unity

[surreal moon]

Same thing

[sturdy crystal]

if i choose a number out of all the real numbers
what is the probability of it being 1

[lethal path]

It is 0 but then that doesn't mean that choosing that number is impossible

A rigorous treatment of this concept would use the concept of a measure

[sturdy crystal]

0 means zero tho

[lethal path]

Yeah

[surreal moon]

That's when you get into uniform distribution

[lethal path]

Well no, you could have a set of the same size (cardinality)
So the set of all integers is countably infinite
The set of all even numbers is countably infinite
But the probability of choosing an even number out of all the integers is 1/2

[lethal path]

If the number of events that you are choosing is finite and the sample space is infinite, then yes

That's not the only way however

The probability of choosing an integer out of the set of real numbers as you said is 0

Cause there are only countably many integers

But there are uncountably many real numbers

[sturdy crystal]

but then how does the zsum of probabilities equal one

@lethal path

[lethal path]

It's an axiom

So we define probability as the probability of all events sums to 1

You can't have more than 100% you know

[sturdy crystal]

but its not one in this case

[lethal path]

Why do you think it's not one?

[sturdy crystal]

adding zeroes cant give one

[lethal path]

It can actually

For example in integration, the area under each line is zero

But adding infinitely many lines together, you get an area

And that area will be a non-zero number

[lethal path]

Your intuition breaks down when you are adding infinitely many things together

This is only true for finitely many things

[sturdy crystal]

infinity is a concept far from the reach of my puny brain?

[lethal path]

So $\lim_{x \to \infty} (1/x + 1/x + \cdots + 1/x) \ne \lim_{x \to \infty} (1/x) + \lim_{x \to \infty} (1/x) + \cdots$

[grand pond]

south

[lethal path]

Yes, great mathematicians spent centuries tackling the concept of infinity

You're not alone

[sturdy crystal]

hmmm

then how do i satisfy my question ❔

[sturdy crystal]

.

[lethal path]

So in your case, $P(E) + P(S \backslash E) = 1$
So that must imply that $P(S \backslash E) = 1$

[grand pond]

south

[lethal path]

Oh right

[lethal path]

Read this again

The answer is no

Cause it depends on the size of the event space

[sturdy crystal]

probability of choosing an odd no = 1/2
and that of choosing an even no = 1/2

[lethal path]

Yes

[sturdy crystal]

this sums up to 1

[lethal path]

No 1/2 + 1/2 = 1

Yeah

[sturdy crystal]

ok

[lethal path]

But we can have infinitely many odd numbers and infinitely many even numbers

[sturdy crystal]

gahh

ill just leave this topic for now

thanks for the help buddy

.close

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[lethal path]

No worries

[midnight plank]

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[last slate]

I just was curious about visualising the geometric effects of having the field, J, be BOTH irrotational and solenoidal here?

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[lime silo]

Task: Solve the following equation over the set of real numbers:
Task: Solve the system of equations over the set of real numbers:
Task: Solve the following inequality over the set of real numbers:
Task: Solve the following inequality over the set of real numbers:
Task: Solve the operation. Provide the result in fractional form (without square roots):

[lime silo]

can someone help?

[midnight plank]

@lime silo Has your question been resolved?

[lime silo]

no

[midnight plank]

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[inland patio]

I have a question about the above definition of a monotone class. There are two definitions; the above one and another, probably more common one, which says that a monotone class is a collection of sets such that it is closed under monotone countable unions and monotone countable intersections. Are they equivalent? I struggle with verifying that they are.

[runic hamlet]

complements essentially turn intersections into unions

so they should be equivalent

[inland patio]

well, complement of a union turns into an intersections of complements. Would you say that the above definition says that M is closed under complements? It says that B \ A only when A < B.

[runic hamlet]

well E is in it

[inland patio]

ok, fair

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[inland patio]

.close

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[icy tundra]

How do I continue here?

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[last slate]

guys is dis right

[last slate]

ping pls

@helper

<@&286206848099549185>

[normal portal]

what does this mean though

[hasty haven]

What is |•|

[last slate]

1.1 means

[last slate]

1.1

[normal portal]

oh

[last slate]

im trying to understand how decimals work and wat they mean

[normal portal]

then yeah 1 + 0.1 = 1.1

basically

[last slate]

mo

noo

the visual

representtion

look st it

[normal portal]

the value of each place is one tenth the value of the following place

[last slate]

is it right

[hasty haven]

doing it with a circle is kinda weird tho

[last slate]

oof ik alalt

[normal portal]

the second on is

[golden wagon]

Well yes with the whole circle

[last slate]

is it right or NOT

[golden wagon]

whole circle + 1/10th of the circle

[last slate]

shutup im unique

[hasty haven]

the area of either is not easily comparable

[last slate]

wdym

ok well

can u guys help

[golden wagon]

As a general representation ig

[normal portal]

ok so after you learn i recommend studying basic vector fields

[last slate]

w the visual representation p,ss

[hasty haven]

yes

[last slate]

im a visual learner

[hasty haven]

the answer is yes

[last slate]

pls guys

[normal portal]

yes its correct

[hasty haven]

u should compare it with a rectangle tho

[last slate]

why not circle

[hasty haven]

at least for me it’s so much easier

[normal portal]

rectangles are easier

[last slate]

coz minr sint perfrct?

[golden wagon]

ppl learn it with pizzas so circle is ig more logical

[last slate]

yea it eas a random ass circle

[hasty haven]

rly

[last slate]

aigh thanks guys

[golden wagon]

idk they teach fractions at school with like pizzas

[normal portal]

you shouldnt be cursing at your age fr

[hasty haven]

ok I dont know the education system anymore ig

[normal portal]

whar

[normal portal]

education system is ass

[golden wagon]

Yep

makes math boring

[normal portal]

wont allow me to take calculus III until like 12th grade

[hasty haven]

jfc

[normal portal]

what

[last slate]

ok guys

thanks

[last slate]

waht

yea ikr

it could be anything tbh

always bout pizzas fr

educational ssystem sucks coz its old asf same sh examples

crammin before exam

makes me forget everything ngl

[hasty haven]

sooooooo dont cram

[last slate]

hate learnin after da

[last slate]

ya

true

[normal portal]

i cant even take Calc I even tho im in grade 8 😭

[last slate]

but anyway

[last slate]

u can self study dawg

no ones stopping u 💀

[normal portal]

im already doing that

im on integration

but i wanna take it

[hasty haven]

see but do they allow u to take AP calc courses

[last slate]

cool

[normal portal]

for fredits

[last slate]

ok bye

im closint his

[normal portal]

only Grade 11

[last slate]

imma close dis

and stap yalls convo

[normal portal]

thy end is now

[last slate]

indeed

.close

[midnight plank]

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[normal portal]

womp womp

[velvet atlas]

very stopped

[last slate]

fsmn why unreact it

[golden wagon]

Grade 8 and on calc wow

[last slate]

anyway

aigh gn yall

[velvet atlas]

goal

[last slate]

ill step on u nibble

alr nvm

bye

[normal portal]

someone im my future hs is on complex analysis

[hasty haven]

jesus

[normal portal]

same grade as me

hes scary

[hasty haven]

in ur grade im pretty sure u have alg 2, trig, discrete calc, limit calc, diff calc, int calc, DE, RA left, and hes on CA

[golden wagon]

Frickin crazy

[midnight plank]

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[full adder]

i got zero*infinity

and do not know how to get rid of it

[surreal moon]

Lhopital or factoring if you are clever

[full adder]

i didnt learn lhopital yet

so i cant use it

how could i factor that?

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[worthy kiln]

in vector fields, fidning the potential function, im sort of confused about the whole + c thing

[worthy kiln]

i dont get how or why we need to do it

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[rough grove]

could anyone help me with this?

[rough grove]

do i just find normal acceleration

and then use newton's second law?

[midnight plank]

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@rough grove Has your question been resolved?

[undone wind]

well

what i can see from this

there should be no vertical net force as there is no acceleration

(vertically)

right?

so the only force would be the one keeping the particle on the circle

aka centripetal net force.

@rough grove u still here?

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[rough grove]

ohh i gotchu

[rough grove]

so the only accleration would be normal acceleration which is v^2/r?

[undone wind]

centripetal acceleration

which is v^2/r yea

and then u need to find v

and there should be a way to find it

in terms of the period and radius jst sayin

distance / time

¯_(ツ)_/¯

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[hot zinc]

i'm not sure how to do this

[final monolith]

A to B is pi/6

[hot zinc]

So what would b - c be?

b to c

[final monolith]

C is halfway between B and D

so b to c is pi/12

you have to memorize that the basic angles are pi/6, pi/4, pi/3 or 30, 45, 60

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[pliant sorrel]

I found the two stationary points

by setting the first derivative to 0

and i got (-1,-6) and (2,-32)

which is wrong

[astral garnet]

u should set the second derivative to 0

[pliant sorrel]

why

everytme i do this its always first derivative

[astral garnet]

well how do u normally solve this?

[pliant sorrel]

find the two stationary points

then just make the inequalities

[astral garnet]

aight

so assuming u know what to do, u just want help with the stationay points?

[pliant sorrel]

wait

no

im suppos to find the infliction points

right

and infliction points is second derivative

[astral garnet]

right

yeah

[pliant sorrel]

and stationary points is first derivative

ok i got it

thanks

.close

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[icy tundra]

Does anyone have a video for solving these

Limits with trigonometric functions

[deep quail]

I am sorry but I can read anything from the image can you elaborate

[deep quail]

Oh yes let me find one for you

[icy tundra]

Because I’ve found this one:

https://youtu.be/HbtuSC_WOW0?si=TIinVWNEI59NCTT4

But I feel like it’s something different

Or like there’s some formulas which I don’t know how to find

[deep quail]

Its for when you have access to calculators I feel

[icy tundra]

I have a calculator

[deep quail]

so where do you face problem

trying to simplify the equation?

[icy tundra]

Like why are the trig. functions squared 💀

[deep quail]

can you give me an example ?

since the image you provided before isn't rendering clear enough for me to actually read

[icy tundra]

Is this better?

[deep quail]

Yes

As you might know we dont have direct formulas for limits of trigonometric functions under square roots and we do need to square it to get the limits in a more simpler form so we can solve it

same with 2nd last one

we don't have a direct formula for limit of 1-cosx^2/x^2

we converted it to sinx^2

as we know that limit of sinx/x

so we can easily split it into two limits as they are in mutiplication

I don't think I can give a more simple reasoning for it you might wanna ask someone else who is good at explaining

[icy tundra]

And could you please help me find the formulas for the trigonometric functions which are not under sq root

Like (cos x-sin x)=1

[midnight plank]

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[midnight plank]

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[icy tundra]

What’s the difference between c) and e)

Why is the solution to e only x-2/x-4

[alpine gyro]

Should’ve substituted 3 in e.

[rain wasp]

did you miswrite e?

the numerator should have been x^2, not x^5

[midnight plank]

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[rustic tusk]

What is the value of x in degrees?

[rain wasp]

well what have you tried?

[rustic tusk]

I thought it was 98 degrees, but I'm not certain

So I asked here to make sure

[rain wasp]

no way x is 98deg lol

[rustic tusk]

Yeah that's what

[rain wasp]

show your work

[rustic tusk]

Oh nvm

I made a mistake

@rain wasp Is it 40?

[rustic tusk]

Angle QOS = 180 - 2(32) = 116

So angle QPS = 180 - (116/2) = 122

x = 180 - (122 + 18) = 40?

.close

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[lunar acorn]

Hey can someone teach me this trick ques ?

[nimble copper]

what's giving you issue?

[lunar acorn]

the question ask for these them solve for x

[queen ermine]

Basically you can make the denominator x

x = 1 + 1/x

[nimble copper]

i know what the question says. I'm asking what you're having issues with

[lunar acorn]

I have no idea how gon start it first

lmao

[fleet moss]

#help-49 message

[lunar acorn]

nope

[lunar acorn]

i cant just substitute x into the ... part ?

or thats meaningless

[queen ermine]

Well it's self referential isn't it?

[lunar acorn]

well yea

[queen ermine]

And then you solve the equation

It simplifies to x^2 - x - 1 = 0

you ignore the negative solution since x must be positive given the initial declaration

and there ya go

Desmos approved

[lunar acorn]

i mean like may i know why ? or which part of the question imply you that x = 1 + 1/x ?

[queen ermine]

Well look at the denominator

It goes on forever

the 1+1/1+1....

so that whole thing down there

is basically x

because x is defined to be 1+1/1+1...

so you can replace that mess with x

[lunar acorn]

ohh i see ....

[midnight plank]

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[sage helm]

$\sum_{i = 1}^{\infty} \sum_{j = 1}^{\infty} \f{i^2j}{3^i(j3^i + i3^j)}$

[grand pond]

uoǝu

[sage helm]

Really not sure how I'd get started on this

$\sum_{i = 1}^{\infty} \f{i^2}{3^i} \sum_{j = 1}^{\infty} \f{1}{3^i + \f{i}{j}3^j}$

[grand pond]

uoǝu

[queen ermine]

have you solved problems like this before?

[sage helm]

Not exactly "like this" before

[queen ermine]

I'd first replace i and j to be a and b for the sake of not confusing myself lol

yeah but like infinite sums

[sage helm]

I have

But not really double summations

The ones I have done usually involve the binomial coefficients

[queen ermine]

Id first try and simplify that right expression

I think you can do some addition on the bottom to get that j to the top

[sage helm]

huh?

[queen ermine]

wait are those i's in the exponent?

[sage helm]

yes

3^i

$\sum_{i = 1}^{\infty} \f{i}{3^i} \sum_{j = 1}^{\infty} \f{1}{\f{3^i}{i} + \f{3^j}{j}}$

[grand pond]

uoǝu

[sage helm]

nice and symmetric

[queen ermine]

well it simplifies to (j) / ((j3^i + i3^j))

[sage helm]

yeah I'm aware

[queen ermine]

wait how'd you do that

[sage helm]

pulled an i out of the inner summation

[deep quail]

Just out of pure curiosity you don't solve this by hand do you?

[sage helm]

you do

[deep quail]

jesus-

[queen ermine]

it's doable

[deep quail]

I have never done these before let me see if i can find anything like this

[queen ermine]

with limits

[sage helm]

wdym by "with limits"

is it a squeeze theorem thing

[queen ermine]

then I guess you're not supposed to use limits, wonderful

[sage helm]

😭 what

I can use anything

I didn't think of squeezing it

[neon rune]

I don't think theres any way of manipulating this into a telescopic sum

[sage helm]

yeah me neither

Now that squeezing is on my mind

I think our best bet is sqeezing the inner sum

and then probably differentiate the geometric series

[queen ermine]

Are you sure this expression is correct?

[deep quail]

what is squeezing are there some under graduate level sums that i don't understand at all?

[sage helm]

Yeah

[sage helm]

It's just using the squeeze theorem

[small jasper]

Have you tried switching the order of summation

[sage helm]

That's beyond the scope of my current level

Let me try some squeezing

[small jasper]

Can you use $$\sum^{\infty}{a=1} \sum^{\infty}{b=1} f(a,b)=\sum^{\infty}{a=1} \sum^{\infty}{b=1} f(b,a)$$

[grand pond]

Civil Service Pigeon

[sage helm]

Probably not

I haven't been taught the theorems regarding allowing switching

[small jasper]

Eh resort to partial fractions then ig

[sage helm]

How do you partial fraction this?

[small jasper]

Let ||a=i3^j, b=j3^i||

[queen ermine]

I tried a numerical calculation, it seems to converge to 0.281, so expect something like that as the final answer

[small jasper]

,calc 9/32

[grand pond]

Result:

0.28125

[sage helm]

Squeezing isn't working

I think I'm on to something though wait

[small jasper]

If squeezing works then show me

I’m curious

[sage helm]

I've discovered something

Very odd

If you let m = n

You get the expected answer

9/32

😭

[small jasper]

psst … it’s because of this

[midnight plank]

@sage helm Has your question been resolved?

[sage helm]

Huh

Is that valid for all summations

[midnight plank]

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[small jasper]

@sage helm As long as the double sum converges absolutely, you can switch the order

[sage helm]

Yeah I did know of that

Now here's something I thought

You wouldn't give me a summation to work on if it doesn't converge

Therefore all switches are justified

I'm sure @robust merlin would agree

[small jasper]

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[tranquil turtle]

how do I factor this

[tacit blade]

…

[tranquil turtle]

$x^2 + (\frac{1-\sqrt{2}}{2})x-\frac{\sqrt{2}}{4}$

[grand pond]

rynite

[tranquil turtle]

sorry I was distracted

I can't think of two numbers that multiply to c and add to b

[tacit blade]

Try completing the square

[tranquil turtle]

can I use the quadratic formula

[slender walrus]

both approaches will work

[tranquil turtle]

oh ok

[tacit blade]

Competing the square is just the quadratic formula reversed

Fun fact

[tranquil turtle]

didn't know

nice

my calculator is giving me decimal values

I want them rational

fractions

[worthy wing]

Start with (x+(1-sqrt2)/4)^2

And work out from there

Its a couple of steps

[tranquil turtle]

what is this?

[worthy wing]

The start to complete the square

It is a way to factorize also

[tranquil turtle]

oh ok ill try

[worthy wing]

Remember one thing

What i did is different from what u have

[tranquil turtle]

alright

[midnight plank]

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[past aurora]

how can I convert this to a general term a_n?
0, 1/2, 0, 1/3, 0, 1/4, …
so I can find limit?