help-36

[rustic trench]

why do we do this

can't i just do u-sub

[scarlet sequoia]

for finding easy primitive

[scarlet sequoia]

so it would work here

[rustic trench]

i could say u = ln(x)

[scarlet sequoia]

go bolder

[rustic trench]

1/(x ln(u)) after the u-sub

[scarlet sequoia]

1/uln(u)

[rustic trench]

i said u = ln(x)

[scarlet sequoia]

yes

[rustic trench]

du = 1/x (dx)

[scarlet sequoia]

yep

[rustic trench]

so we have 1/ln(u) du?
where du = 1/x dx and u = ln(x)

[scarlet sequoia]

?

[signal vector]

Write it on a piece of paper

You might be confused if you do it in your mind

[scarlet sequoia]

that would have been for...

[rustic trench]

$$
\int \frac{1}{x \ln (\ln (x))} d x
$$

Substitute $u=\ln (x) \longrightarrow d u=\frac{1}{x} d x$:
$$
=\int \frac{1}{\ln (u)} d u
$$

[soft zealot]

dgh

[scarlet sequoia]

this integral

[signal vector]

We have 1/(xln(x)ln(ln(x)))

[scarlet sequoia]

unfortunately the reason I told you to multiply by 1/ln(x) is because 1/ln(u) is indeterminate (at least no simple primitive

[rustic trench]

i've not understood the reason we would do 1/ln(x) multiplication yet

[scarlet sequoia]

you know what let's stick from here

do we agree that ln(u) < u?

[rustic trench]

yes

because e^ln(u) < e^u

[scarlet sequoia]

so 1/ln(u) > 1/u

[rustic trench]

okay

[scarlet sequoia]

so this integral $\int \frac{1}{\ln (u)} d u$ is bigger than...

[soft zealot]

rafilou2003

[rustic trench]

int(1/u)

[scarlet sequoia]

yes

which is

[rustic trench]

ln(u)

[scarlet sequoia]

and what can we say about ln(u) when u gets to infinity?

[rustic trench]

it goes to infinity

[scarlet sequoia]

uh huh

[scarlet sequoia]

and so this integral?

[rustic trench]

goes to infinity?

[scarlet sequoia]

diverges

[rustic trench]

or i mean, depends on the endpoints, it's not 0 atleast

[scarlet sequoia]

so this sum?

[scarlet sequoia]

one of the end points is 30

[rustic trench]

diverges?

[scarlet sequoia]

the other is infinity

[scarlet sequoia]

by integral comparison test

[rustic trench]

the integral comparison test part was?

[scarlet sequoia]

$\sum f(n)$ has same convergence type as $\int f(x)dx$ when $f$ is decreasing

[soft zealot]

rafilou2003

[scarlet sequoia]

(and integrable)

[rustic trench]

okay and f is decreasing because the denumerator is increasing -> inverse is decreasing?

[scarlet sequoia]

yes

[rustic trench]

the part that confuses me is we're talking about sums in the first part

adding something means we add something so it becomes bigger

but how is it similar to an integral which kind of calculates infinitely small rectangular areas?

[scarlet sequoia]

the proof is here if you want to check

an integral is an infinitesimal sum of those rectangles

[rustic trench]

okay so to summarize how the problem was solved

\begin{enumerate}
\item Ensure the function (f(x) = \frac{1}{x \ln(\ln(x))}) is continuous, positive, and decreasing for all (x \geq 30). These conditions are necessary for the integral test to be applicable.
\item Use the integral test, comparing the series to the integral of its continuous counterpart. Thus, we set up the integral (\int_{30}^{\infty} \frac{1}{x \ln(\ln(x))} , dx).
\item The integral (\int_{30}^{\infty} \frac{1}{x \ln(\ln(x))} , dx) was calculated. The result of this integral indicates the behavior of the series.
\item If the integral is finite (converges), then the series also converges. If the integral is infinite (diverges), then the series diverges.
\end{enumerate}

[soft zealot]

dgh

[rustic trench]

Did I skip over any important step, or get something wrong?

@scarlet sequoia

[scarlet sequoia]

that's it

[rustic trench]

or technically we didn't do step 3

[final saddle]

@rustic trench Has your question been resolved?

[rustic trench]

@scarlet sequoia got time for this?

also i have a follow-up question if you got time

ping me when you see this

[final saddle]

@rustic trench Has your question been resolved?

[scarlet sequoia]

we didn't do step 3 because we computed something smaller that went to infinity

we've shown that $\int_{30}^{\infty} \frac{1}{x \ln(\ln(x))} , dx \geq \infty$

[soft zealot]

rafilou2003

[scarlet sequoia]

so it's infinite

by comparing it with $\int_{ln(30)}^\infty \frac{1}{u}du$

[soft zealot]

rafilou2003

[rustic trench]

how is something geq infinity, infinity doesn't ever end? actually i don't recall how we've shown that function goes to infinity, isn't the function technically just decreasing?

[rustic trench]

and i don't recall this part, when did we change the endpoint with ln(30)?

[scarlet sequoia]

well if it's equal to infinity

[scarlet sequoia]

the function integrates to ln(u)

[scarlet sequoia]

u sub

the bounds change

and when you get $[\ln(u)]^{u\to \infty}{u=\ln(30)} = \lim{u\to \infty}\ln(u) - \ln(\ln 30)$

[soft zealot]

rafilou2003

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@rustic trench Has your question been resolved?

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[supple dove]

Hello I need help solving this problem

so I’ve started by trying to use induction on n

the induction step is really simple

What’s hard is proving the base case

I know I can argue that if o(t(0))=o(o(0)) then by the peano axioms that implies t(0)=o(0)

I’ve also thought about left inverses since I know t and o are injective by peano axioms

however I still can’t seem to get anywhere with this proof

Also sigma ( or o ) here represents the usual successor function

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[crisp epoch]

for the inequality 2cosx(cosx - 1) > 0, the only thing we need from the (cosx - 1) part is it's critical value because cosx > 1 is never true right?

[tiny gorge]

it's never true, so if the inequality holds, that means you have 2cos(x) times a negative number is positive

which tells you what about 2cos(x)?

[crisp epoch]

oh thats a very nice way of explaining this situation

[crisp epoch]

it has to be negative

[tiny gorge]

yep!

[crisp epoch]

thanks

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[warm python]

How does $-1<cos(u)<1$ imply $cos(1)<cos(cos(u))<1$?

[soft zealot]

Why am. I here

[warm python]

I mean I know why this is true

but how does statement 1 imply statement 2 is true

[rocky tusk]

take the cos of the everything for the inequality

cos is an even function

[warm python]

ah

yeah, makes sense now

thanks

[rocky tusk]

ur welcome

[warm python]

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[tranquil pine]

need to determine the derivative

[tranquil pine]

I used product rule here because it made sense

but I dont' seem to be on track

$(\tfrac12 x ^{-\tfrac12})(x^3-x)+(3x^2-1)(x^{\tfrac12})$

[soft zealot]

wyldinwilliam

[lyric summit]

so far ok

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

<@&286206848099549185>

[lyric summit]

it looks very well

[lyric summit]

finally you can write this way:

[tranquil pine]

why is it written like this instead

your right but

I'm just curious why this format is more simplified

$\frac{\sqrt{x}(7x^2-3)}{2}$

than this ^^

[soft zealot]

wyldinwilliam

[lyric summit]

i offeerd you form as your teachers wants, since i am too. both are equivallent

it looks just nicer

[tranquil pine]

👍

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[final saddle]

Please don't occupy multiple help channels.

[sage river]

Uh

I wasnt??

i got kicked out of the other one...

[shadow aspen]

#help-24

[sage river]

Everytime i leave this channel it disappears

why is that happening

[shadow aspen]

Click on MATH HELP (OCCUPIED)

[sage river]

o

[tulip coyote]

You probably have the help channels hidden - make sure that it's expanded so you can see read ones

[sage river]

oh i see

but anyways

i dont get this

like f(x) isnt vertex form so what kind of form is it

like is it just partial factoring thing

[shadow aspen]

try expanding ax(x-s)+t

what do u get

[shadow aspen]

yeah pretty much

[sage river]

um

[shadow aspen]

factor first 2 terms ignoring last term

[sage river]

ax^2 - axs + t

[shadow aspen]

yeah that is in the form ax^2+bx+c, so just do that in reverse

starting from ax^2+bx+c

[sage river]

ok so theres 4 forms

standard form, vertex, factor, and partially factored?

because im confused on the difference of factored form and partially factored

[shadow aspen]

This isnt a commonly used form, but you could say that

fully factored is like a(x-b)(x-c) while partially would be what is shown above

[sage river]

right

alr

ty

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[glad basalt]

can someone explain what is going on here

How does multiplying y cause that

[coral pagoda]

he plugged in y (prolly given in the beginning?)

[glad basalt]

oh right

lmaoo

what about this

derivative of f(x)

what rule is happening here

[coral pagoda]

properties of ln

[glad basalt]

😭

[glad basalt]

solvable like this you think

[coral pagoda]

uhhh

i would just say use ln properies, its more convenient cuz that approach may take longer to simplify

well more messy i should say

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[half pumice]

Stuck on how to apply the order limit theorem to get to the last line from the previous. There doesn’t seem to be a good way to compare the two series so I’m thinking I have to apply part (iii).

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@half pumice Has your question been resolved?

[half pumice]

<@&286206848099549185>

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@half pumice Has your question been resolved?

[edgy bane]

maybe this is naive, but can’t you construct a sequence c_n where c_i=c for all i. then clearly it converges to c, and since c_n = c \leq b_n for all n then we can apply (ii)

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[mellow jacinth]

could someone help me with this question for part b?

[final saddle]

@mellow jacinth Has your question been resolved?

[mellow jacinth]

<@&286206848099549185>

[errant night]

Use vietas theorem

Formula

[mellow jacinth]

i havent learnt that yet

[errant night]

Oh. Then you have to do it the hard way

[mellow jacinth]

i know the sum and product rule

[errant night]

I am gonna solve and see what happens

Gimme 2 mins

[mellow jacinth]

i already did part a i just need to do the part b

ayt

[errant night]

Nvm,

Not able to solve

[mellow jacinth]

okok

[errant night]

Sorry

[mellow jacinth]

its aight dw

[unreal anvil]

let r =ratio of the GS
use abc=8
show that ar=2

[mellow jacinth]

ay thank you

[mellow jacinth]

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[thin coral]

how do i isolate y

[white tiger]

complete the square for y

[thin coral]

ahh

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[golden ledge]

The angle of depression from the top of the mountain to the horizon is 2 degrees. The angle of elevation from viewpoint number 1 and viewpoint number 2 to the top of the mountain is 10 degrees and 20 degrees respectively, and the distance between the two viewpoints is 12 km. What is the height of the mountain and the true radius of the Earth? There is another mountain roughly with the same height 5000 km away on the ground. Give the arc distance between these two mountains and give the straight-line distance between these two mountains.

[golden ledge]

ive found the mountain height, 4.1km

but i dont quite understand how to find the radius

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@golden ledge Has your question been resolved?

[golden ledge]

<@&286206848099549185>

[boreal rock]

can u like draw so i can understand

[golden ledge]

im not sure how I can draw the problem out

im super confused on this part

[boreal rock]

same its just if we draw we could understand more

i hate wordings

[golden ledge]

i think its something like this but

it dosent seem right?

[boreal rock]

hm i dont think so

[golden ledge]

from what i understand, theres a person looking from a mountain 4.1km high and hes looking down 2 degrees and he can see the horizon?

[boreal rock]

ill try draw it

[golden ledge]

alright, thanks

[golden ledge]

ive run the question through ai, and it from what ive seen its not possible to calculate the radius. I think I should just ask my teacher about this question tommorow.

[boreal rock]

maybe the question is wrong who knows

[golden ledge]

yea probably, thanks for trying though!

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[jade fable]

Please inform me if there is anything that seems obscure.

I’m trying to find the t.

Then I found that angle(ADB) is 90 degrees

[jade fable]

Which contradicts to the circumstance that lineDA and lineAC intersect with one another.

Any idea?

Is there anything wrong?

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@jade fable Has your question been resolved?

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[smoky plover]

hello

[smoky plover]

if $f$ is not piecewise continuous over $I\implies \int_I f$ diverge

[soft zealot]

Adam Chebil

[smoky plover]

can you give a counter-example to this ? (other than the popcorn function)

[desert mantle]

why not that one

[smoky plover]

it's so complicated, i couldn't even tell if it's piecewise continuous or not

[desert mantle]

well its not

thats the point

any function which is not piecewise continuous will be complicated like that

the popcorn function is a relatively simple counterexample all things considered

[barren hound]

i mean you could define a function with only one "pop"

[desert mantle]

well that would be piecewise cont

[smoky plover]

wait, piecewise continuous according to the french definition of piecewise continuous right ?

i think i showed it to u

[desert mantle]

bla bla can be split into intervals on which it is continuous

[smoky plover]

on which it is piecewise continuous

[desert mantle]

no on the individual intervals it is continuous

[smoky plover]

ln(x) for example is piecewise continuous on ]0, infty[

[barren hound]

would $f(x) = \begin{cases} 1 & x = 0 \ 0 & \text{otherwise} \end{cases}$ be piecewise continuous?

[soft zealot]

hayley!

[desert mantle]

its also just continuous

piecewise continuous is like the word basically says, it can be split into pieces and on each of these pieces it is continuous

every continuous function is continuous because you can split it into one piece

[desert mantle]

I think under any reasonable def it should be

[smoky plover]

this is not piecewise continuous

[desert mantle]

you shouldnt care about the jumps

[smoky plover]

you should

[desert mantle]

no otherwise floor(x) for example also isnt piecewise cont

if you want that it is continuous at the jumps then you just have a normal continuous function

[smoky plover]

can you understand french ?

[desert mantle]

no

[barren hound]

non pas

[barren hound]

there's no way you can split it into intervals that are continuous bc it's not continuous at 0

[desert mantle]

(-infty, 0) and (0, infty)

[smoky plover]

i think this is piecewise continuous

[desert mantle]

and who cares about jumps

[smoky plover]

the limits must be finite (not necessarly equal)

[barren hound]

you're just leaving 0 out in the dust?

[desert mantle]

I would treat it as a "partition" and then say it needs to be cont on each open interval

which iirc is what adams book does

(for bounded domains)

[smoky plover]

yeah according to the definition, f is piecewise continuous iff f is continous on any open interval included in I

[bronze dock]

oh

[desert mantle]

the whole intuitive point behind piecewise cont functions is that they are basically continuous with maybe some jump points in between

[smoky plover]

and the limits at the bounds of every interval must be finite

@desert mantle so according to that definition, the popcorn function is not piecewise continuous and the integral converges ?

[desert mantle]

no

the popcorn function is not continuous on any interval

[smoky plover]

oh i meant not piecewise continous srry

[desert mantle]

yes

[smoky plover]

ok thnx

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[pine ridge]

how was this simplified

[analog gale]

Consider the denominator first 2^2 = 4 and 3*4 = 12, next up with the numerator, club the like terms (4-h)^2 multiplied by (4-h) is (4-h)^3 ... Pi remains as it is.

[pine ridge]

oh i see

thx alo

t

[analog gale]

@pine ridge You can close the question if you're done, ig

[pine ridge]

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[tranquil pine]

How to prove triangle △MNE is similiar to △ECD

[tranquil pine]

i am stuck

all i know is ∠NEM = ∠CED

[analog gale]

I suppose MN and CD are parallel as well?

[tranquil pine]

ahm

wait

[analog gale]

Only then, you can argue some other angles as equal as well ... Do you see how?

[tranquil pine]

Segment MN is the midline of triangle ABC. Line MD intersects the continuation of side AC at point D, and side BC at point E (see figure).
1.What fraction of the area of triangle ABC is the area of quadrilateral AMNC?
2.Prove that Triangle MNE ~ triangle DCE.

full question

i have solved the 1.

[tranquil pine]

I guess

[analog gale]

Yeah, it is a theorem, the midline MN is parallel to the base AC

and hence it is parallel to the continuation as well

[tranquil pine]

Alright

how do i get other angles?

i only need 1 of each

[analog gale]

Well, if MN and CD are parallel, angle MNC is equal to angle NCD, no?

[tranquil pine]

bruh

ofc

how i dont see that

do you know how those angles are called?

so i can do more research on that

I dont really know all the rules, most of the time i guess by the eye

[analog gale]

It's just parallel lines cut by a transversal, ig ... which makes a buncha angles equal

Sorry, I have been outta school for quite a long time, I don't remember the exact terminology

[tranquil pine]

Alr alr

ill find it

thx alot tho

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[analog gale]

No, don't do this

[tranquil pine]

😄

[analog gale]

I suggest, go read up on both the above facts I mentioned

first how the midline is parallel to the base

second, how are angles in parallel lines equal

Get an intuition as to why this is correct, and remember them for a lifetime ... Hope that helps

[tranquil pine]

Yes it will thx

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[final saddle]

@steady blaze Has your question been resolved?

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[long kraken]

triangle bisector which goes through 9cm and 6cm sides splits third side with 2 sections and one of the sections is same size as one of the side. Find third side of the triangle

[long kraken]

I think this question can have couple of answers but in the answers page it only had 1 answer which was 10 so how can I solve it so I will have only 1 answer?

[final saddle]

@long kraken Has your question been resolved?

[long kraken]

nope

I still need help

[candid pulsar]

well what method would you take then

like do you have a plan on how to solve it

[long kraken]

I started with the bisector rule a/b = a1/b1 => 9/6 = (I took 9 as the section) so 9/6 = 9 /x

and then solve for x

and 9 + x = third side

9x = 54

x = 54/9

x = 6

but 6 and 9 is 15 as the third side but answer was 10

[candid pulsar]

but just wait

now what are the sides of the triangle

[long kraken]

9 and 6

[candid pulsar]

9, 6, and 15

[long kraken]

yeah

[candid pulsar]

but is such a triangle possible?

think about it

[long kraken]

9 and 6 mnust be more then 15 right? dd

[candid pulsar]

yep

so it won't be a triangle

[long kraken]

I think it could be equal to 15 too no?

[candid pulsar]

but itd just be a line :/

9 and 6 on one side and 15 on the other

[long kraken]

so the strategy would be to check all the possible ways and then the one that is the triangle would be the answer?

[candid pulsar]

well what do you mean by using 9 as the section here

[long kraken]

bisector splits third side as 2 sections and one section is equal to one of the sides

[candid pulsar]

maybe, but we'll check your method and if it works, then we'll work from that

[candid pulsar]

ah

[long kraken]

so I chose 9 as one of the section

[candid pulsar]

but 9 can be in the denominator too

there are two sections right?

[long kraken]

yeah

[long kraken]

so that's why I am confused that there is only 1 answer

[candid pulsar]

so another possible equation is 9/6 = x/9

[long kraken]

in the answers page

[long kraken]

yep exactly

[candid pulsar]

yeah but maybe some cases don't work

[long kraken]

yeah that must be the reason

or maybe I don't give correct condition of that question

[candid pulsar]

hmm we'll see if this works then

[long kraken]

it's on Georgian language so I am not sure if it's correct

[long kraken]

alright

so if it can't be triangle

then 2 of the equations are gone

[candid pulsar]

the case fails

yeah

[long kraken]

2 of them led to 15

[candid pulsar]

so either 9 in denominator or 6 in numerator

for 9 in denominator we get x = 81/6

which is 13.5

so the third side is

[long kraken]

in that case c is 22.5

[candid pulsar]

yep

[long kraken]

but answer was 10

[candid pulsar]

but again

[long kraken]

😄

[candid pulsar]

it fails

[long kraken]

that's so confusing

[candid pulsar]

since 9 + 6 < 22.5

so it's not possible

only case left is 6 in numerator

[long kraken]

wait why? biggest side is oh yeaah

big side should be smaller

then other 2

[candid pulsar]

yeah

[long kraken]

ohh that must be the solution elimination of other cases

[candid pulsar]

now finally we have 9/6 = 6/x

and guess what x is

[long kraken]

4

[candid pulsar]

yes

[long kraken]

but there is one issue

[candid pulsar]

so the third side is 6 + 4

[long kraken]

I found out just now

[candid pulsar]

oh?

what is it?

[long kraken]

here is the problem

we can't assume that 15 is not the answer

it could be equal

to third side

[candid pulsar]

greater than or equal to?

huh

[long kraken]

yeah

[candid pulsar]

thats weird

so 15 or 10?

[long kraken]

yeah but confusing part is that answer was only 10

in the answer's page

so that's so strange

[candid pulsar]

idk does the question talk about something like the triangle shouldn't be a line or something

[long kraken]

maybe there is some issue that we don't include

[long kraken]

no only the condition I told you

[candid pulsar]

since the 9 6 15 triangle is basically a line

oh

what

but all of these say greater only

[final saddle]

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[candid pulsar]

bruh

[long kraken]

but the screenshot I gave you was from wikipedia

and I also heard that before

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[long kraken]

another one

this one

[wet warren]

It must be only greater than the third side
If it is equal to the 3rd side then it is just a straight line not a triangle

[long kraken]

why are there some sources that say it should be greater or equal?

I guess that's the answer

I just don't get why

is that

if it's straight line then it can't be called triangle

[wet warren]

Maybe that they consider any connected 3 lines as a triangle but as I said it is just a straight line that has a point in the middle if the sum of 2 sides equal to the third side

[long kraken]

I kinda get it if we had just a line and 3 points and 1 segment is equal to 2 other

maybe that's what they consider as triangle?

[wet warren]

Yeah

[long kraken]

ok thx for help ❤️

.close

[final saddle]

Channel closed

Closed by @long kraken

Use .reopen if this was a mistake.

[long kraken]

what am I closing btw? :d

😄

it doesn't looks closed to me

.close

.close

[barren hound]

it will actually lock here in a sec

[long kraken]

kk

[barren hound]

it gives you some time to reopen if you made a mistake

[long kraken]

oh ok

let it free up then

[barren hound]

it will :)

[long kraken]

.reopen

[final saddle]

✅

[long kraken]

real qick

quick

I have 1 question

😄

[wet warren]

Just ask the question

[long kraken]

if we have paralelogram and from the obtuse angle we want to allow 2 heights how can we do that?

1 is from the angle immediately

to the top

but another one?

where can it be?

do you want me to give you photo of what I mean?

[wet warren]

If you really need to get the height out of the obtuse angle you can extend the line above it then draw the height

[long kraken]

I am doing it on bottom obtuse angle so 1 height would be to do on the top and another one on the right side for example?

well I will give you question of that and you might understand better

[long kraken]

from paralelogram's obtuse angle's vertex there are 2 height's 2cm and 3cm and angle betweenthem is 30degree find diagonals

[long kraken]

that's the condition

so how would I do 2 heigts

from 1 angle

?

would it work?

those 2 height

s

but they form 30degree angle and I don't think it looks like 30 degree

😄

cuz it's not

[final saddle]

@long kraken Has your question been resolved?

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@long kraken Has your question been resolved?

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[past sparrow]

I need someone to help me with thease:

[past sparrow]

Let f be a cubic function.

For each of the statements below, decide whether it is true or false. Justify the answer.

Claim 1: The graph of f has at least one extremal point.

Claim 2: All lines of the form y = ax + b, where a,b ∈ R, will intersect the graph of f.

Claim 3: If the graph of f has an inflection point for x = 3, then f'(1) = f'(5).

[final saddle]

@past sparrow Has your question been resolved?

[final saddle]

@past sparrow Has your question been resolved?

[final saddle]

@past sparrow Has your question been resolved?

[past sparrow]

.close

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[plucky fern]

Financial Maths question:
a) Using the given information, estimate the value of the bond portfolio if the interest rate is raised to 4.6%.

b) Suppose that the investor invests his capital into this bond portfolio for 5 years and the interest rate becomes 4% in coming 5 years, estimate the IRR of this investment.

I have calculated the answer for the first part to be 472.36, I am a bit confused about part b)

[sly marten]

we have nothing to work with
what is "the given information"?

[plucky fern]

this is what I got at the moment

I am not sure if the 4.6% interest rate is related to part(b) at all

[final saddle]

@plucky fern Has your question been resolved?

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@plucky fern Has your question been resolved?

[plucky fern]

.close

[final saddle]

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[faint berry]

Can someone tell me how I solve for the problems in the box?

[final saddle]

@faint berry Has your question been resolved?

[faint berry]

.close

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[vale python]

how to prove this set has no limit points?

it diverges, so theres no limit point the set itself converges to, but how do you prove systematically that none of the infinitely many elements in the set are not limit points?

is it like a proof by induction somehow?

[rain compass]

consider the difference between any two terms

you'll see that it is a nonzero finite amount. so if, say, the sum up to 1/N were to be a limit point, you'd need a sequence which converges to the sum up to 1/N

[neon slate]

it's really easy to see with integral test also btw if you are familiar with that, then you just need to consider
$\int_1^\infty\frac{dx}{x}$

[soft zealot]

Soosh

[rain compass]

for eps < 1/(N+1), there will be no such sequence which eventually gets within epsilon of the sum to 1/N

[neon slate]

which is ln|x| evaluated from 0 to inf and clearly has no limit

[vale python]

we've not covered the integral test yet

[neon slate]

it's pretty easy to get the idea of it just reading the definition above if you wanna peek ahead

[rain compass]

soosh im not sure why you brought the integral test in

[vale python]

yeah it does seem pretty clear, but I know my prof wont want me to use it on homework

[rain compass]

they are talking about the set having no limit points, not that the series diverges

[neon slate]

oh i should read better

[vale python]

is your profile pic a strokes album cover?

[rain compass]

it is yeah

[vale python]

nice

knew i recognized it

hmmm okay though, i think i kind of follow your chain of logic

[rain compass]

to be explicit about what i said above, you'll note that the closest term to a_n in magnitude is a_{n+1}

[vale python]

right

[rain compass]

and that difference is exactly 1/(n+1)

um btw a_n here is the sum to n of 1/i

but anyways

if you wanted to construct a sequence that converged to a_N, you'd need to find some elements that eventually satisfy
|a_N - a_n| < 1/(n+1)

but that is impossible because of what we mentioned above

[vale python]

ohhh i see

[lyric summit]

you can also notice that no element of your set is the limit of any subsequence of a given sequence, and this is because the given series diverges to infinity

[vale python]

yeah that also sounds like a familiar theorem

that might even be easier

[lyric summit]

it comes from metric toplogy

[vale python]

thanks all for the help!

.close

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[vale python]

hello, im back

[vale python]

i realized that showing that each term of this set are not limit points is not sufficient to show that the set is closed.

i am lost as how to formally prove that this is a closed set.

[shadow aspen]

Are there any limit points?

[vale python]

no

there are no limit points

[scarlet sequoia]

meaning that if a sequence is convergent...

[vale python]

but how can i show there are no limit points?

i put isolated points up there but i meant none of the terms are limit points

[scarlet sequoia]

ok, here's a different approach

find the complement set

show it's a union of open intervals

[vale python]

okay, so that would be like (-infty, 1) U (1, 1+1/2)U(1+1/2,1+1/2+1/3)U...

oh thats way easier

is there a more clever way to write that complement though that im not seeing?

[scarlet sequoia]

no that's exactly how you're supposed to write it

[vale python]

wow

ive been working on this problem for a bit and now its just going to be like 3 minutes to write up lol

thank you so much

[scarlet sequoia]

if you want to go the alternate way of showing any sequence that converges has limit in this set

take any convergent sequence (xn) with values in this set, name it E
name the limit of that sequence x
you know x >= 1
show that E intersected with [0,2x] is closed
conclude that x is in that set, so in E

[vale python]

im not sure i follow

[scarlet sequoia]

you know that a set E is closed $\Longleftrightarrow$ for any convergent sequence $(x_n){n\in \bN}$ with values in $E$, the limit $\lim{n\to\infty} x_n \in E$

[soft zealot]

rafilou2003

[vale python]

right

[scarlet sequoia]

so

you wanna prove the latter statement

"Let a convergent sequence $(x_n)_{n\in \bN}$ with values in $E$"

[soft zealot]

rafilou2003

[scarlet sequoia]

name $x = \lim_{n\to\infty} x_n$

[soft zealot]

rafilou2003

[scarlet sequoia]

Step 1: show $x\geq 1$

[soft zealot]

rafilou2003

[scarlet sequoia]

Step 2: show $E\cap [0,2x]$ is closed

[soft zealot]

rafilou2003

[vale python]

why [0,2x]?

[scarlet sequoia]

[0,2x] is a closed set that is a neighborhood of x

any other set like that works

[vale python]

okay

that makes sense

[scarlet sequoia]

Step 3: find a sequence that converges to $x$ with values in $E\cap [0,2x]$

[soft zealot]

rafilou2003

[scarlet sequoia]

Step 4: deduce that $x\in E\cap [0,2x] \subseteq E$

[soft zealot]

rafilou2003

[scarlet sequoia]

End

Do any of those steps seem too difficult for you to attempt on your own?

[vale python]

hmm okay

no i dont think so

maybe step 3 is a little fuzzy

[scarlet sequoia]

basically use epsilon definition of convergence with epsilon = x

[vale python]

okay yea thats what I thought

well thank you very much!

.close

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[peak hemlock]

Hi! I'm struggling with some surjectivity and bijectivity proofs.

My current problem is:

if $f: A -> A$ and $f(f(x))$ is bijective, is $f(x)$ bijective? Either write a proof, or find a counterexample.

Note: I am NOT looking for an answer. I would like guidance.

I've tried for about 3 hours in total now with a few approaches.

[soft zealot]

Nameless

[peak hemlock]

My current thought is that $f$ must be bijective.

If we assume it is not surjective, then $f(A) \subsetneq A$. By the definition of a function, the image of a function under a set $X$ cannot have a higher cardinality than $X$. Consequently, it's impossible for $f(f(A))$ to be onto $A$, since $f(A) \subsetneq A$ and $f$ must therefore be subjective.

[soft zealot]

Nameless

[peak hemlock]

I'm not sure if that's correct, and also don't know how to prove injectivity

[daring hound]

The definition of injective is one-to-one

That is, if you have f(a)=f(b), then a=b

[peak hemlock]

I love discord. What the hell.

[daring hound]

Block him…

[peak hemlock]

Done already

[daring hound]

Yes

So since the domain and codomain are both A in this case

[peak hemlock]

First, is my reasoning for the onto-ness of f correct?

[daring hound]

Yes

[peak hemlock]

Perfect. Thanks ~

[daring hound]

I was trying to show the concepts to you, but looks like you are pretty good with these already

[peak hemlock]

Yeah. I think I understand the concepts. I just struggle with proof writing

[daring hound]

Hmmm

Try this

This should destroy this question in no time

So f(f(x)) is definitely included in the example above

[peak hemlock]

in this, is gf(x) function multiplcation or compsition ?

[fading dock]

I'm not sure if you even need this proposition. If $f(a) = f(b)$ then certainly $f(f(a)) = f(f(b))$. But since $f(f(x))$ is bijective, then it is injective so we have that $a = b$. Right?

[soft zealot]

Awesam

[peak hemlock]

I had that $f(f(x_1)) \neq f(f(x_2))$ when $x_1, x_2 \in A$ and $x_1 \neq x_2$.
If $f(x_1) = y_1, f(x_2) = y_2$, and $y_1, y_2 \in A$, then we have

$$f(f(x_1)) = f(y_1) \neq f(y_2) = f(f(y_2))$$

I decided this was wrong since we don't know that $f(x) = y$ where $x \in A$ necessarily outputs all $y \in A$. The knowledge this holds for all $y$ is part of the definition of injectivity.

[soft zealot]

Nameless

[peak hemlock]

This seems like simmilar reasoning to you, @fading dock ?>

OH! Since I have shown the surjectivity of $f$, all $y$ are necessarily outputted. Consequently, this holds for all $y_1, y_2 \in A$ where $y_1 \neq y_2$, thus showing injectivity and therefore bijectivity of $f$ ??

[soft zealot]

Nameless

[fading dock]

yeah I think that reasoning holds after you have shown surjectivity as you figured out. it's kind of similar to what I said just coming at it from the opposite direction

[peak hemlock]

I see. You're showing that $f(a) = f(b)$ for all $a, b \in A$ where $a = b$, where I'm showing the not-equal version of the same thing

[soft zealot]

Nameless

[peak hemlock]

yours seems significantly simplier though

[fading dock]

yeah you just showed the contrapositive of what I showed

also I didn't exactly show that $f(a) = f(b)$ whenever $a = b$. what I showed is that $f(a) = f(b) \implies a = b$, which is logically equivalent to $a \neq b \implies f(a) \neq f(b)$, which is the definition of injectivity

[peak hemlock]

Thanks both of you. I appreciate the help

[soft zealot]

Awesam

[peak hemlock]

I see! I've been operating on a slightly too strict version of injectivity, then. I conflated $f(a) = f(b) \implies a = b$ with $$f(a) = f(b) \iff a = b$$

[soft zealot]

Nameless

[peak hemlock]

wait. i was wrong in the above message.
a = b \implies f(a) = f(b) is trivial and holds for all functions.
I'll have a think about what my confusion actually was, then.
thanks again.

.close

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[bright current]

Im a bit stuck on this

[lusty pecan]

could you explain what part? it seems you have it pretty much solved

[bright current]

I understand a bit since thats what we've been doing all week but im stuck on is minimum and the range

[lusty pecan]

ah alright

[bright current]

I know maximum is 1 and -4

but is it the same for minimum?

[lusty pecan]

well just think about it graphically

does the parabola ever stop

[bright current]

No its infinite

or -oo,oo

[lusty pecan]

so does it have a specific minimum

[bright current]

no

So DNE?

[lusty pecan]

and there's your answer

[bright current]

for both of minimum?

[lusty pecan]

yeah

because there's no x-value for which the minimum exists

[bright current]

Ok so 1 and DNE

But how do we figure out the range?

[lusty pecan]

what y-values does the graph exist on?

[bright current]

1?

Thats all I get when I plug it into desmos

[lusty pecan]

like

you understand what range is?

[bright current]

Yes

The Y-axis

[lusty pecan]

so, looking at the graph, where does the function exist and where does it not exist with respect to y-coordinates

[bright current]

1 and oo?

[lusty pecan]

I think you get it but it's negative infinity to 1

[bright current]

Oh because it starts from the left side I see so like this (-oo,1)?

[lusty pecan]

we're referring to y here so we think about it from down to up instead of left to right

so when you start from the bottom

that's negative infinity as you established

and then it stops at 1

[bright current]

Oh you mean

(-oo,1]?

[lusty pecan]

yes

[bright current]

Uh wait I just submitted it

And I missed something

Is not 1?

[lusty pecan]

there's no minimum

its -infinity but the minimum is looking for a value so it's just DNE

[bright current]

So DNE as well?

I thought that was just for the X-value?

[lusty pecan]

well it's like

you're plotting a point without an x-value

does the point exist

[bright current]

Oh so you would have no where to start and no where go

No

it doesn't

[lusty pecan]

yep

so it doesn't exist at all

[bright current]

Got it but I can ask we do one more to see if i got it?

[lusty pecan]

sure

[bright current]

One second

Ok so this one is new to me

But for A

would it be (0,1)u(6,oo)?

[lusty pecan]

yeah

[bright current]

ok and now for B

I see it decreases at (4,6)

[lusty pecan]

yes

[bright current]

Awesome

so now where its constant

Is it (1,4)?

[lusty pecan]

yeah

[bright current]

Uh ok D stumped me here

Because the line is running across the X-axis

[lusty pecan]

does that matter here

[bright current]

Oh so -oo,oo?

[lusty pecan]

yep

[bright current]

ah ok

as for the range

Im not sure but I have [4,oo)

[lusty pecan]

think about it from down to up

what values does it exist on?

y-values

[bright current]

-3

and

4?

[lusty pecan]

does it stop at 4?

does it exist at let's say y=5

[bright current]

Im not sure I don't think so because its not intersecting with it

would it be oo?

Are you still there?

I figured it out but one of them I got wrong

[warped rivet]

Yes

[bright current]

Oh hi

[warped rivet]

Oh

there’s more than one interval

[bright current]

ok so like this?

(-oo,-3)u(1,4)

?

[warped rivet]

Yeah

Sait

Wait

[bright current]

oh I didn't put anything in yet

[warped rivet]

Why -3?

we’re dealing with x-coordinates when we talk about intervals

[bright current]

Oh!

ok so (-oo,0)u(1,4)

[warped rivet]

Yep

[bright current]

Im not sure thats right

ah awesome

Arightyo I might be back later on but thank you 😄

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[lime vapor]

can someone help me with 3

[lime vapor]

heres where im at rn idk if its correct

<@&286206848099549185>

[honest tartan]

i love you

but i can’t help

i’m flirting with a girl

this is a one time moment for me

[lime vapor]

💀

[hard elbow]

I would first combine the denominator so you just have a fraction on a fraction. Then use the rule where you multiply and flip the second fraction.

[lime vapor]

would this be correct?

[hard elbow]

correct so far

[lime vapor]

what would i do next?

[hard elbow]

Divide both fractions

[lime vapor]

?

wat fractions?

[hard elbow]

You have 5/(x+4) divided by (3x+8)/[x(x+4)]

divide the fractions

Then simplify

[lime vapor]

im confuse how would i divide them?

[hard elbow]

Do you know how to divide two fractions?

[lime vapor]

like this?

[hard elbow]

Yes

and then that turns into a multiplication sign and the second fraction is reciprocated

[hard elbow]

Thats it

[lime vapor]

would i do this

[hard elbow]

Now expand and simplify

[hard elbow]

yes you can do that

[lime vapor]

would this be the final answer?

[hard elbow]

Yes, that is correct

good job

[lime vapor]

how could i simply this more

[hard elbow]

First step is to put the fraction together

[lime vapor]

how would i do that?

would i mutiply the right side by 6 and left side by 2x-7

[hard elbow]

you multiply x by 10

and 6 by 2x-7

you always multiply across

[lime vapor]

so 10x/12x-42

would that be the final answer

wait i could divid by 2

[lime vapor]

so would 5x/6x-21 be the final answer?

[hard elbow]

yes

[lime vapor]

alright theres a rly hard 1 and ima need alot of help

[hard elbow]

I do not understand this method.

[lime vapor]

ight wat would u do

[hard elbow]

I would simply combine fraction in the denominator

then simplify the fractions individually in both numerator and denominator

then divide the fractions with the multiplication method

[lime vapor]

could u help me find the denominator

[lime vapor]

.

[hard elbow]

Sorry mate im super busy right now

[lime vapor]

alright im just remake the post

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[reef patio]

Is there a way I can learn Maths faster for my upcoming exams?

[silver maple]

There is no shortcut. practice textbook questions with fully worked solutions, followed by past papers

[reef patio]

are there any resources you can recommend?

[silver maple]

You have not specified year level or topic

[reef patio]

IGCSE Additional Mathematics

The topics that are a challenging are circular measure, combinations and permutations, trigonometry

[molten remnant]

pnc is all about practice

trigonometry is learning formulas

now idk what is circular measure

[reef patio]

everything about circles (sectors, segments, archs, etc.)

[silver maple]

I am not familiar with IGCSE. I am familiar with HSC. I think you might be interested Terry Lee. https://advancedmathematics.com.au/products.htm. For your case due to syllabus changes it will probably be best to go libgen and see if you can find an old textbook that covers circular geometry

[reef patio]

Alright thanks

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[final saddle]

Please don't occupy multiple help channels.

[winter kernel]

this isn't right but im not sure what im doing wrong

[fading dock]

you found the critical points but one of them corresponds to a local minimum and the other corresponds to a local maximum

you need to identify which is the minimum

[winter kernel]

ohhhh im illiterate 50% of the time ty

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[open hazel]

I am graphing systems of linear inequalities to find the solutions.
I run into a problem where i get the correct lines but when filling in. (you fill in the area to the graph that satisfies both inqeualities)

My question is :

When you change inequality into a temporary equation.
If during the simplification of the equation you have to multiply the equation by II *(-1)

do you change the >,< behind the scenes?

example:

x-y < -5

x - y = -5 II -x
-y = -5 - x II * (-1)
y = 5 + x

I can gather the info to the line from here but when testing witch side of the line gets filled in the test is:

Substitute x,y with a point (0,0)

y < 5 +x

0 < 5+ 0

0 < 5 TRUE

Or should i have changed the < from the original inequation into > since we multiplied by -1?

[white tiger]

you can do the simplification with the inequality itself

x - y < -5

x + 5 < y

y > x + 5

but yes if you multiply -1 you need to change the inequality sign still

[open hazel]

Okay! So in a way i could do the whole process without changing the inequality into an equation and then i would keep better track of the state of : <,> ?

[white tiger]

yes

[open hazel]

you just use <> as the =

mm okay! I Will do that thanks a lot for the help!

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[floral compass]

Continuous or not

I cant figure out the first step

[west berry]

Just put limit x tends to 1

If x>1 then it'll be the right hand limit

If it's <1 it's the left hand limit

All of them should be equal to what the function equals at x=1 to be continuous

[final saddle]

@floral compass Has your question been resolved?

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[warm elbow]

Hi can someone help with this? Looks impossible without negation operators

[pliant shore]

Start by turning for all into there exists, and there exists into for all

[warm elbow]

Ok I have done that

It’s like swapping the for all x into the there exist for x ??

[pliant shore]

Yes

[warm elbow]

Is that what you mean?

Yeah ok..but how does that negate?

[pliant shore]

If you want to disprove something is true for all x, you just need a counterexample

You need only one example to show that the claim is false for one particular x

[warm elbow]

Ok so I have now on paper “for all y being an element of the real number, there exist a x being an element for the natural number set”

[proud jolt]

$\neg \forall x : p(x) \equiv \exists x : \neg p(x)$

[soft zealot]

tales

[warm elbow]

What does this mean?

[proud jolt]

'-'

[warm elbow]

Ok that’s true

[proud jolt]

That's the negation of forall x

[warm elbow]

This????

[pliant shore]

In plain English, the negation of "for all x, P is true" is "for (at least) one x, not P is true"

There exists an x where P isn't true

[warm elbow]

Does that involve using conditional or negation operators? The question said I can’t use any of that

[proud jolt]

It will simplify on the end

for example

[pliant shore]

Well if you want to negate xyz > 0

That's just $xyz \le 0$

[soft zealot]

south

[proud jolt]

Yeah, like south said

[pliant shore]

Stuff like that ye

[warm elbow]

I’m lost..so like right now; I’m here.

What do I do next?

[proud jolt]

.

[warm elbow]

The equivalent part on the right is the statement I need and it doesn’t involve the use of conditional and negation operators?

Ohhhh

I get it

Hang on

What about the z part? “There exists a z part of the rational number set whereby xyz >0”?

Do I just xyz<0 and it will suffice?

[proud jolt]

<=

[warm elbow]

Or do I need to change the statement in front as well?

[proud jolt]

You need

[warm elbow]

I change it to “ for all z part of the rational family, there exist xyz <= 0”?

Like that?

Is that right?

[proud jolt]

seems correct