help-39

[hazy radish]

f: Z -> Z is bijective, but is not the identity function f(x) = x

if it came from a finite set it would be obvious

but this is anything but

[calm wing]

any linear function is bijective

(except from the constant function ofc)

for example f(x) = x + 1 is bijective

f(x) = 2x is bijective etc.

[hazy radish]

but how do you know that?

[west sapphire]

f(x) = 2x is not bijective on Z

[calm wing]

oh yeah mb

but x + n is

[west sapphire]

yep

f(x) = -x works too

[calm wing]

the inverse of x+1 exists and is x-1

every integer is sent to another unique integer

[hazy radish]

heres what my confusion with that is. Cant you prove bijection if the cardinality of domain and codomain is the same?

[west sapphire]

that proves that a bijection exists (that's the definition of cardinality), but it doesn't help you construct one

[hazy radish]

true

its just confusing thinking of an infinite set

[calm wing]

does the above clear up your doubts?

as said, having an example problem would be very helpful

[hazy radish]

sorta?how does it prove bijection

[calm wing]

by "it" you mean two sets having the same cardinality?

[hazy radish]

no, im assuming the above you meant proving the function and the inverse exists

[calm wing]

oh, well

there's this thing
https://en.wikipedia.org/wiki/Schröder–Bernstein_theorem

[calm wing]

but also non-bijective functions don't have proper inverses

think about it: suppose f is NOT injective, that is, there are distinct a and b s.t. f(a)=f(b)

let f(a)=f(b)=k, so what is f^-1(k) equal to? a or b? the inverse is not well defined

similar argument goes for (non) surjective functions

[hazy radish]

oh, cause if the codomain is a constant there is no possible way to find the inverse

[calm wing]

precisely

[hazy radish]

ah

[calm wing]

think of a bijection as uniquely "renaming" each element

[hazy radish]

thats also why its important to have the same cardinality for domain and codomain

and why every element in the domain should be tied to a unique value in the codomain

that makes sense

[calm wing]

https://math.stackexchange.com/a/1215724/1153468 nice answer formally proving this'

[calm wing]

yep

[hazy radish]

how do you know if a function has an inverse?

prove its surjective and injective first

im assuming

but then how would you prove that?

for an infinite set

[calm wing]

again, it would be easier if you had an example

[calm wing]

if it's bijective it has an inverse

[hazy radish]

how about something like f(x) = x^2

[calm wing]

is it injective?

[hazy radish]

yes

[calm wing]

you sure?

[hazy radish]

it should have the same domain and codomain

[calm wing]

f(3) = f(-3) but 3 =/= -3

it's not injective

[hazy radish]

oh

thats right

[calm wing]

injectivity can be tested with a horizontal line test

[hazy radish]

so the codomain is smaller

[calm wing]

er

not quite

[calm wing]

if the domain and codomain are the same cardinality then there exists SOME function that is bijective

[hazy radish]

true, i know we set both

[calm wing]

however this is not really constructive

[hazy radish]

or it might be the range, i confuse the two

[calm wing]

it's just some function has to exist

[calm wing]

codomain in this case

[hazy radish]

gotcha

but it does make sense why its not injective

elements in domain dont have unique values in the codomain

or something like that

[calm wing]

yes

[hazy radish]

i feel im messing up the thinking

so it also wouldnt be surjective depending on what you set the codomain to be

[calm wing]

you really gotta get well familiar with the terms and i know it takes a while cuz i still sometimes mess them up

[hazy radish]

like it wouldnt be if the codomain is all integers

[calm wing]

yes

or all reals, for that matter

[hazy radish]

right

so how do you come up with functions that meet specific properties?

[calm wing]

why? there is no k s.t. f(k)=-1

[hazy radish]

right

I believe the largest codomain you can have for it to be surjective is natural numbers

[calm wing]

not quite

non-negative reals works too

[hazy radish]

yes, mb, i was thinking of discrete values

[calm wing]

yeah okay

[calm wing]

it's a very broad question. usually there are some maps that just make sense why they would be bijective. idk. i've stared at desmos for too long and it just clicked lol

[hazy radish]

true, its pretty easy for finite sets cause you can just create the sets that meet the certain criteria

its just hard thinking of infinite sets

cause you cant create an infinite set

[calm wing]

what makes finite sets easier for you?

[hazy radish]

you can set the cardinality

[calm wing]

suppose you need to find a bijection between {1,2,3,4,5} and {1,2,3,4,5}, how would you do it?

(that isn't the identity function)

[hazy radish]

well it would have to be surjective so every element in the codomain is used, and every element in the domain has a unique element in the codomain.
Off the top of my head f: {(1,2), (2,3), (3,4),(4,5),(5,1)}

[calm wing]

great!

this function can be described by a simple rule f(x) = x+1 and wrap around if x=5

drop the "wrap around" part and you have f(x) = x+1, a bijection on Z

[hazy radish]

oh

i didnt even think about that

ya, that would make sense if it was an infinite set starting at the value 1

o anything actually

[calm wing]

this is your finite bijection visualized

(at least, that's how i personally imagine it)

[calm wing]

extending that function to the entirety of Z makes it look like this

two uneding infinite strips - but all connections are very much unique and nothing is left out

[hazy radish]

thats exactly how i visualized both

[calm wing]

awesome, then the intuition for infinities and stuff will come to you very soon

[hazy radish]

sweet, so just to make sure, iforgot who mentioned, all linear functions are bijective?

wait nvm

[calm wing]

not on Z

[hazy radish]

we covered that x^2 isnt

right

[calm wing]

x^2 is not linear either

[hazy radish]

thats right

[calm wing]

by linear i meant mx+b and stuff

[hazy radish]

exactrly

x^2 is polynomial

[calm wing]

any more doubts left?

[hazy radish]

no, this has defintely helped. Nowi just need to practice

thank you for the help

.close

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[floral cedar]

Hi there, I'm learning about infinite sequences and series in Calculus II. I'm trying to solve a problem where I must test an alternating series to see if it converges. To do so, I'm using the alternating series test, where I need to perform two checks:

1. Check that (b_{n+1} \leq b_n ) for all (n)
2. Check that ( \lim_{n\to\infty} b_n = 0 )

I'm a little stuck on step 1, where I must ensure that each subsequent term in my series is smaller than the previous one. My series looks like this:

[
\Sigma_{n = 1}^{\infty} \frac{2n(-1)^n}{n+4}
]

Thus, my (b_n+1) and (b_n) are:

[
b_{n+1} = \frac{2(n+1)}{(n+1)+4} = \frac{2n + 2}{n+5}
]
[
b_{n} = \frac{2n}{n+4}
]

Now, it isn't obvious at first sight which of these are larger than the other. I've thought of reducing (b_{n+1}) to the following form:

[
b_{n+1} = \frac{2n + 2}{n+5} = \frac{2n}{n+5} + \frac{2}{n+5}
]

And now it's clear that the first term in the sum is smaller than (b_n) for all n, and it's also clear that the second term in the sum is smaller than (b_n), but I'm not sure how to prove that their sum is smaller than (b_n) for all n.

Can anyone give this a look? Thank you!

[jolly parrot]

de Beauvoir

[summer imp]

Try and do the limit first. It's easier and will save you some trouble

[floral cedar]

Alright, I'll do the limit.

For my own reference though, is there a way to check which sequence is larger?

It seems to me that if we use the decomposed version of (b_{n+1}), as (n\to\infty), the second term will approach zero. So for very large values of (n), we should only have to compare (\frac{2n}{n+4}) with (\frac{2n}{n+5})

[jolly parrot]

de Beauvoir

[floral cedar]

And now it looks like b_n+1 is smaller, since it has a larger denominator.

[summer imp]

In some instances, unless it's really clear which is bigger, it can be insightful to compute $b_{n+1} - b_{n}$ and see its sign.

[jolly parrot]

Azyrashacorki

[floral cedar]

Oh, I see.

That makes much more sense than trying to take limits, I guess.

[summer imp]

But yeah I think in general, taking the limit of the term is the easiest step and fairly quick, so if it doesn't go to 0, you can immediately stop.

[floral cedar]

It's also clear that my intuition is wrong, since if we graph out the two terms, it is clear that b_n+1 is larger

(Here the blue graph is b_n, and the green and purple graphs are the first ratio and the second ratio in the sum that makes up b_n+1. Their sum, while not graphed, is clearly larger)

[floral cedar]

Anyways. I thank you for your help

You're right that the most straightforward approach to these types of problems is to just test the limit of b_n first, before doing all this weird mumble jumbo..

I'll proceed with that approach and also remember to compute b_n - b_n+1 as per your suggestion.

[summer imp]

Yeah, that helps in most cases, and it's easier to check for the sign than to guess which is bigger.

[floral cedar]

Oh, and on the topic of checking the sign

ChatGPT offers another approach, where we check the inequality through algebraic operations across the inequality

Is this approach also valid?

[summer imp]

That works, but it's more or less equivalent in steps to taking b_{n+1} - b_{n}, using common denominators and investigating the sign.

So yeah it's fine

[floral cedar]

Alright, gotcha.

Thank you very much for your kind and gentle help tonight, @summer imp

I am very grateful for your consideration.

[summer imp]

No problem, feel free to ask more questions whenever

[floral cedar]

Just used your method and took the difference between b_n and b_n+1. Got a fraction that reads -8 / (n+4)(n+5)

This means b_n+1 is indeed larger than b_n

It checks out! 🌈

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[jovial hare]

they never mentioned rationalizing a denominator so what would be more simplified?

[steep marlin]

wdym more simplifed

[rough stream]

"most simplified" is personal preference

Sometimes it's obvious which the preference should be. In this case, I think they're pretty similar

[steep marlin]

i would just rationalize since most teachers want rationlized answers

[maiden badger]

in trig they always rationalize, when it comes to sin, cos and tan of 30 45 and 60

[floral cedar]

I personally like the first one, because reciprocals are cute! 😍

[jovial hare]

so its the image on the right?

root10/10

[steep marlin]

yeah

always rationalize

[sly dagger]

generally this is better

[steep marlin]

unless the question says not to

[pearl pond]

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[sly dagger]

can someone explain why arg(z)=arg[z-(1+i)] on the argand diagram looks like this?

[gusty flume]

think about what arg(z) means and what you are plotting?

you are plotting each point on the complex plane for which that given equation holds

arg() can be interpreted as describing the "direction" of the arrow from the origin to the complex number in question

for which complex numbers does subtracting 1 and then subtracting i not change that direction?

[sly dagger]

so if i were to pick these points as z and z-(1+i)

[gusty flume]

correct

[sly dagger]

ah k

[gusty flume]

now try picking a point not on the diagonal where Re(z)=Im(z)

and then try picking a point that is on the diagonal, but specifically in the gap left there

[sly dagger]

yeah i see it now

thanks

do u think u are able to explain questions that involve arg[(z-z1/(z-z2)] = theta?

[gusty flume]

what's z1, z2?

[sly dagger]

like 2 random points

[gusty flume]

consider what z-z1 or z-z2 are

then what you get by dividing them

and at that point the argument is again the "direction" of that, which is supposed to be some constant theta

[sly dagger]

hmm i dont really get it, it forms some arc

[gusty flume]

try interpreting it step-by-step, from the inside out

[sly dagger]

well these 2 are lines from z1 to z and z2 to z

[gusty flume]

okay next step

[sly dagger]

that should subtract their angles

or rather arg(z-z1)-arg(z-z2)= theta

[gusty flume]

so in more imprecise language, this plots all complex numbers z, s.t. the line from z1 to z and the line from z2 to z stand at a certain angle to on another

[sly dagger]

right, i sort of see how that would form an arc, but i dont at the same time

[gusty flume]

thale's theorem is a specific case of this

I think this is just the inscribed angle theorem expressed in complex coordinates instead of 2D Euclidean space?

[sly dagger]

never heard of this

[gusty flume]

https://en.wikipedia.org/wiki/Thales's_theorem

In this example, A and C are your z1, z2

this would be the case where theta=90°

then all points z (B in this graphic) satisfying your equation lie on an "arc" (perfect semicircle in the case of 90° )

[sly dagger]

right

[gusty flume]

the inscribed angle theorem is a more general case and basically states that no matter what secant you draw, the angle formed at B remains constant (though not necessarily 90°, unless you pick exactly the diameter)

[sly dagger]

hmm ok

[gusty flume]

so you'll always get a "partial circle" of sorts when plotting this

[sly dagger]

if theta<90 then we'd get something like this right?

pretend that this is a fantasticc circle

why dont we see a mirror image of this on the bottom?

[gusty flume]

because we're dealing with oriented angles that care about direction

ie arg(z-z1) - arg(z-z2) is not equal to arg(z-z2)-arg(z-z1)

you'd get the lower side by either subtracting or adding pi radians (180°) or multiplying with -1

[sly dagger]

so if we had arg(z-z2)-arg(z-z1) this would be flipped?

[gusty flume]

I would expect so, yes

[sly dagger]

hmm ok

thanks again

ill try to understand this more

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[warm roost]

how did we use change of base to get the rhs in the circled equation

[vital estuary]

inside is log(log(x))
base is a^2

[versed remnant]

u could let "log_a^2 log_a^2(x)" for "u" for a bit so u have
log_a^2 (u) , and then change the bases
[ not needed but helps understand]

[vital estuary]

log_a(inside)/log_a(base) is what you get

[warm roost]

where did u come from

huh?

[vital estuary]

its a substitution

[warm roost]

how did u get this

[vital estuary]

imagine replacing the inside of the first log_a^2 with a box

[warm roost]

a box?

is that a mathematical term?

[vital estuary]

no

substitution is the mathematical term

basically you stuff it inside a box to deal with later

[warm roost]

what is being substitued by "u"

[vital estuary]

you get log_a^2(□)

log_a^2(log_a^2(x)) is the □

[versed remnant]

i made it more confusing for him

[warm roost]

yes

[warm roost]

okay

[vital estuary]

just treat the box like any variable

[warm roost]

does that help alot?

[vital estuary]

can you see how to change the base for log_a^2(□)?

[vital estuary]

yes, this helps

[warm roost]

yes

then?

[vital estuary]

how can you change log_a^2(□) to something base a?

[warm roost]

change of base?

[vital estuary]

yes

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[inner timber]

If we have a strong choquet space does it mean that it has no points which are isolated ?

[inner timber]

or i havent understand it correctly ?

[pearl pond]

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@inner timber Has your question been resolved?

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@inner timber Has your question been resolved?

[pearl mist]

<@&286206848099549185>

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@inner timber Has your question been resolved?

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[midnight haven]

guys question

[midnight haven]

for this would u use the straight line or the line in which u r looking at the tree

cause sometimes u use the ground

sometimes the line above its weird

[open ivy]

none of them, you are supposed to find this length. And they have given the height of the tree and an angle, so I would use those (height and angle) to find the searched length.

[midnight haven]

ye

that's what i meant

where would i use x to label

like the ground or the sight

but yeah u answered it ty

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[molten quartz]

Hi guys

[stuck wasp]

hello

[viscid dew]

please post your problem

[molten quartz]

I'm dealing here with telescopic series

11/6 is the answer

In my lecture books i have this theory

the telescopic series converges to b sub 1 - the limit of the b sub n term

[delicate ember]

For telescoping series questions it's often very helpful to just write out the expanded form

If the summand was $\frac{1}{n} - \frac{1}{n-1}$ you would be right

[jolly parrot]

sed

[molten quartz]

yeah i see most people expand the series

looks like my theory doesn't work for all the telescopic series

[delicate ember]

Yea it also avoids silly shifting errors

[molten quartz]

i have to found the first and last term in the expanded form right?

[delicate ember]

I'd expand it and see what I get

[molten quartz]

i expand it and i got 11/6

[delicate ember]

That's your answer well done!

[molten quartz]

quite tedious having to expand the series

[molten quartz]

i'll investigate why this formula doesn't work sometimes

thanks

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[still pebble]

How to factor x^2+2x-12?

[still pebble]

Do i use the -b +- root b^2-4ab/2*a?

[worthy lance]

Have you factored before? There are a lot of methods

[still pebble]

Damn

[still pebble]

I only know the usual for example since idk what this factorization is called x^2-3x+2 = (x-2) (x-1)

[worthy lance]

Ok so i will show you a mew way

You can use this whenever u want

It is called complete the square

U know this?

@still pebble

[still pebble]

Nope

[worthy lance]

Ok lets start with an easy one

x^2+2x+1

Do you know this one?

[still pebble]

(X+1)^2

[worthy lance]

Perfect

Now the second thing i have to know is

Do you know what is difference of squares equal to?

For example x^2 - 2^2?

[still pebble]

No

[worthy lance]

Ok so this one is equal to (x+2)(x-2)

If you expand u will see it is true

[still pebble]

Oh thats what you meant by difference, the pos neg?

[worthy lance]

Ok so now lets apply this two thing to our exercise

Yes

[still pebble]

Aight

[worthy lance]

We have x^2+2x-12

We know that x^2+2x+1=(x+1)^2 right?

[still pebble]

Yes

[worthy lance]

But we dont have + 1

Lets write it

x^2 + 2x + 1 -12

Now

You can see this is not equal right?

[still pebble]

Yes

[worthy lance]

But what about this x^2 + 2x + 1 -12 -1

Is it equal now?

We added 1 but subtracted 1 too

So it is like adding 0

[still pebble]

Yeah

[worthy lance]

Ok now we take the part we know

(x+1)^2 -13

U follow me?

[still pebble]

Okay yeah

[worthy lance]

Now lets rewrite the 13

13 = (sqrt(13))^2

Do u agree?

[still pebble]

Yeah

[worthy lance]

Ok so we have

(x+1)^2 - (sqrt(13))^2

Lets make this clearer for you

Let’s call x+1 = a

And sqrt(13) = b

So we have a^2 - b^2

Do u follow me?

[still pebble]

Yes

[worthy lance]

Factor this

[still pebble]

(A+b) (a-b)

[worthy lance]

Perfect, now

Replace back

[worthy lance]

We know what is a and b

[still pebble]

(X+1 + sqrt(13)) (x+1 - sqrt(13))

[worthy lance]

Finished

[still pebble]

Hoyshit

Nice

[worthy lance]

This is the most satisfying method for me

You can factor with this method even really hard problems

But this is only for quadratic

That is why it is called complete the square

[still pebble]

It was pretty simplistic

Thank you 👍👍

[pearl pond]

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[minor eagle]

so for the last 3 i got:

H. 3/5
I. 1
J. and 1/7

[minor eagle]

but it says they are wrong so idk what to do

nvm i got it, turns out i just needed to use ":"

.close

[pearl pond]

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[minor eagle]

.reopen

[pearl pond]

✅

[minor eagle]

ok so, I did this problem multiple times and i get 3/10 and 3:10

but the website says its wrong

i dont understand what im doing wrong

[midnight haven]

maybe it's a site issue

try 12/52

[quiet tendon]

doesn't the ace count?

[midnight haven]

and 12:40

[quiet tendon]

4 picture cards a suit

and if yiur answer for c is right thrn clearly a is 4/13

[minor eagle]

ok lemme see

[minor eagle]

ok 4/13 is correct

joked and put 4:10 in B but thats wrong

[quiet tendon]

if the probability is 4/13, how would the odds be 4:10?

the odds would be 4:10 if the probability was 4/14

[minor eagle]

read the first word of the sentence

[quiet tendon]

oh i thought you were asking why that was wrong

[minor eagle]

somethings wrong here

B is very tricky

ok lemme try this

I got 4:9

does that seem correct to you

.close

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[minor eagle]

!.close

!close

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[north oriole]

Is negative 45 on the unit circle equivalent to 7pi/4 ?

[viscid dew]

do you mean -45 degrees?

[north oriole]

Yes

[viscid dew]

yes, they are equivalent.

[north oriole]

Ok

[viscid dew]

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[north oriole]

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[rich token]

I’m confused how the integral of 0.1e^-0.002x dx is -50 e^-0.002x

The e stays the same

[sly dagger]

the integral of u' e^u is just e^u

[rich token]

Yes

Wait

[sly dagger]

u need the derivative out the front

[rich token]

Yeah I know it’s u substitution

But

-0.002 is the deravative…?

[sly dagger]

yes

[rich token]

How does the 0.1 go to -50…?

[sly dagger]

0.1/-0.002 i assume

[rich token]

1/-0.002 maybe

Um

Ah I see

You just distributed it back into the integral

👍

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[visual canyon]

How can I determine the type of conic section represented by an equation in general form?

[cosmic charm]

general form is something like ax² + by²+...?

[visual canyon]

wait what

[visual canyon]

ook firstly i dont fully understand the question

my context messages didnt send

for conic sections stuff like

if its equal to zero its a parabola

[cosmic charm]

is it an equation like this

[visual canyon]

mhm

[cosmic charm]

complete the square seperately for x and y

[visual canyon]

could i just look at = 0?

[cosmic charm]

?

[visual canyon]

and determine that way

[cosmic charm]

i dont get how you look at = 0 and determine it

[visual canyon]

im confusing myself too i cant find the thing im talking about

one sec

If A and C are non zero and equal, and both have the same sign, then it will be a circle.
If A and C are non zero and unequal, and have the same sign, then it will be an ellipse.
If A or C is zero, then it will be a parabola.

[unborn abyss]

just complete the square

[cosmic charm]

yeah

i aint remembering that lol

[visual canyon]

so to do this i just.. complete the square?

thats it?

[unborn abyss]

you'd complete the square and then match it to either an ellipse, a parabola, or a hyperbola

(in the case of a parabola you won't be able to complete the square twice)

[visual canyon]

oh

yeah thats it then

thanks!

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[barren root]

What do the small symbols to the right and down mean?
https://wikimedia.org/api/rest_v1/media/math/render/svg/c3ba9a37e86e54a62beb226f2949a840deff8765

[sterile tusk]

evaluated at

so like the first one would be the derivative of z wrt x evaluated at x

[barren root]

The derivative formula is the same everywhere, what does it matter which point we evaluate at?

[sterile tusk]

dz/dx gives us a function that describes the instantaneous rate of change at a given point

that rate doesnt necessarily have to be the same everywhere

lets f(x)=2x, f'(x)=2

the rate of change is the same everywhere

but if we take f(x)=x^2, f'(x)=2x

itll give us a different number depending on what x is

[barren root]

I know the definition of derivative through limits and have worked with derivatives before. I just thought that we would get the derivative formula instead of instantaneous rate of change at a specific point. Does "evaluated at x" mean that x is the horizontal axis when we have "dx" in the divisor?

[sterile tusk]

okay so the derivative formula is $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

[jolly parrot]

y0shi

[barren root]

Yes

[sterile tusk]

its really telling us what the rate of change is between two points as they get infinitely close

but this rate of change is not guaranteed to be the same everywhere

[barren root]

Right

[sterile tusk]

thats why when it says evaluated at x, it means that we are finding what the instantaneous rate of change is at the specific point when x=x

we just use the variable x to describe all the possible values we can plug in there

x could be 4 and itll mean evaluating the derivative at 4

[barren root]

Okay, thanks

[sterile tusk]

yw!

[barren root]

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[visual canyon]

If the solution to a linear system of equations is x=6 and y= 2 and you graphed the two equations, where would the two lines intersect? How do you know?

[visual canyon]

I'm not sure where to start here

[sly dagger]

the solution to a linear system is their point of intersection

[visual canyon]

oh

is that it??

[visual canyon]

the lines would intersect at point (6,2)?

[sly dagger]

correct

[visual canyon]

oh damn that wasn't hard

thanks!

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[restive briar]

How do I get the lower bound here?

[restive briar]

The solution is $x\in(-1,4]$

[jolly parrot]

Luca M

[summer imp]

You have to do some casework

If x+1 >=0, then yes, you get x<= 4.

But if x+1 <= 0, the inequality flips, you'll get some other bound

[inland ivy]

you can't cross multiply inequalities

[restive briar]

ah i see

[astral rapids]

you have to multiply both sides by the square of the bottom and then solve the quadratic

[pearl pond]

@restive briar Has your question been resolved?

[pearl pond]

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[dapper raft]

can anyone helpo me with this?

[pearl pond]

@dapper raft Has your question been resolved?

[open ivy]

May this hopefully help you (one way is finding unshaded BCX and unshaded in ABXD separately and then summing them up).

[pearl pond]

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[fluid relic]

How to find the domain of y=√ tanx and y= x/cosx

[fluid relic]

<@&286206848099549185>

aaa

<@&286206848099549185>

[verbal whale]

$$y = \sqrt{\tan(x)}$$

[jolly parrot]

Alberto Z.

[verbal whale]

This?

[fluid relic]

yes

[verbal whale]

Alright, which between tan and sqrt have problems (i.e. require specific conditions on the input)? Tan, sqrt or both?

[fluid relic]

wot

[verbal whale]

Ok, let's see it this way. What values of "stuff" are in the domain of $\sqrt{\text{stuff}}$?

[jolly parrot]

Alberto Z.

[fluid relic]

tan(x)

its find the domain of that

sqrt of tan x

[verbal whale]

Yeah I understood it 😅

I don't understand why you don't understand my question. Maybe it's a language barrier? Because I'm not English so perhaps I used strange words?

[fluid relic]

yeah i dont understand what you are asking

[verbal whale]

What's the domain of √x ?

[fluid relic]

anything bigger than 0?

[verbal whale]

≥ 0

Also 0 is ok, because √0 = 0

[fluid relic]

wait equal to?

ohh

okay andd

[verbal whale]

And now can you answer to this?

[fluid relic]

tan(x) bigger and equal to 0

yeah but i dont know how

[verbal whale]

Perfect

[verbal whale]

You mean you don't know how to solve tan(x) ≥ 0?

[fluid relic]

yeah i havent done math in a while 😅

[verbal whale]

Have you ever seen trigonometric equations?

[fluid relic]

more like i havent done trig in a while

cuz after ap test

[verbal whale]

Like sin(x) = 1/2

[fluid relic]

yea i know that but tan

forgot

[verbal whale]

Go and revise that, it would take too long for me to explain to you how to solve that

[fluid relic]

;-

;

thats why im here but okay

[verbal whale]

Do you know that tan(x) = sinx/cosx?

[fluid relic]

yes

[verbal whale]

Nice

[fluid relic]

do i must use unit circle

[verbal whale]

There are several ways, I don't know what you're used to

[fluid relic]

wait so

its anything but

pi and 2pi

?

s

[verbal whale]

Nope

What's tan(π/2)?

And what's tan(2π/3)?

And also tan(π) is 0, so it must be accepted, not rejected

[pearl pond]

@fluid relic Has your question been resolved?

[fluid relic]

wth

[fluid relic]

oh yeah

its the pi/2 place

where its undefined

ohhh

oHH

but how do you

notate it

[pearl pond]

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[fluid relic]

.reopen

[pearl pond]

✅

[viscid dew]

reopen for what?

[viscid dew]

^

[viscid dew]

DO NOT REPEATEDLY POST YOUR QUESTION IN OCCUPIED CHANNELS

[fluid relic]

i didnt finish my question

[fluid relic]

.

[viscid dew]

oh sorry

$\tan\frac\pi2$ DNE

[jolly parrot]

Your LaTeX Helper

[viscid dew]

(does not exist)

[fluid relic]

wut

it exists

[viscid dew]

it doesnt

[pearl pond]

@fluid relic Has your question been resolved?

[viscid dew]

If that answers your question, please type .close to allow other people to use the help channels.

[pearl pond]

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[mental verge]

hi, how can i write sin3xcos5x as a sum of trigonometric functions?

[mental verge]

I don't know where to start.

and i've also got a similar question to this

[gentle lichen]

sin2a = 2sinacosa

Split up sin3xcos5x into sin3xcos3x and cos2x and then you can apply that identity

[mental verge]

i thought of doing that but is cos 3x * cos2x = cos5x?

also in this one how can i apply sin2a = 2sinacosa?

doesnt make sense

we got sin3x

[gentle lichen]

By splitting it into those terms, I meant addition of those terms

Wait nvm

I made a stupid mistake

[vital estuary]

there is a trig product to sum formula but i could never remember it

[gentle lichen]

💀

[gentle lichen]

Do you remember the identity of sin3x and cos 5x

Both of those are some long polynomials of sinx and cosx respectively

Then you would have to multiply them

It would be very tedious

[mental verge]

no

[gentle lichen]

But you'll get an answer

[mental verge]

so im just gonna multiply them?

[vital estuary]

when you mean "sum of trigonometric formulas", do you want of the form some sum asin(x)cos(x)?

[gentle lichen]

Sin3x = sin(x + 2x)
This will give you sin 3x
Similarly you can get cos5x using 2x and 3x

[vital estuary]

or would this suffice?

[mental verge]

tbh idk this is the part im confused

like its what the question says

[vital estuary]

!original

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[mental verge]

refer to the highlighted parts

[mental verge]

i did give u the original problem though

[gentle lichen]

Do you have the answer

[vital estuary]

ok yeah you want to use sum to product and product to sum

[mental verge]

i need to check wait

[mental verge]

how did u get that?

[gentle lichen]

Because the wording is not clear

[mental verge]

yeah ik

wait a min

[vital estuary]

theyre rare trig identities

[mental verge]

11

1/2 (sin8x-sin2x)

it says

[vital estuary]

confirmed, use product to sum

[mental verge]

alr

thanks

btw

[mental verge]

am i gonna use the same for the next question as well

12

i highlighted it as well

[mental verge]

.

[gentle lichen]

You would need sum or differece to product identities for that one

[vital estuary]

use this one

[mental verge]

alright thank y'all so much

the wording is so weird that ik how to solve these tbh but i get confused

it could've said solve sin3xcos5x ....

anyways solved both of them correctly, have a great day y'all

.close

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[pearl pond]

Please don't occupy multiple help channels.

[patent badger]

i need help with

Question 2

sEcond pic is what I got

[pearl pond]

@patent badger Has your question been resolved?

[patent badger]

<@&286206848099549185>

[stable basin]

i mean cos(x+pi)=-cosx if you want to simplify, but it looks right

[patent badger]

the criteria says must have correct period, amplitude, phase and centre all correct

is it correct?

[stable basin]

they produce the same thing, yours is fine unless its some online grading system that might screw you over

short ans: yeah, you good

[patent badger]

they mark it irl

[patent badger]

the graph i sent you is good including formula?

[stable basin]

ye

[patent badger]

thank u bro

[pearl pond]

@patent badger Has your question been resolved?

[pearl pond]

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[exotic ledge]

hello, i wanted to ask if there are good free resources on the internett about arithmetic and logic? i want to research the foundation of proofwriting, logic and arithmetic.

[pearl pond]

@exotic ledge Has your question been resolved?

[pearl pond]

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[visual canyon]

Angle α is drawn on the unit circle in standard position. The terminal side of angle α intersects the unit circle at point (-0.292, 0.956). What is the approximate decimal value of tan(α)? how do i do this?

[spare lark]

Tan(x) = sin(x)/cos(x)

Where cos is the x and sin is the y

So sin(0.956) / cos(-0.292)

[visual canyon]

wait hold oon

terminal side of an angle

intersects a circle at a specific point

cos iz x sin is y

why?

[pearl pond]

@visual canyon Has your question been resolved?

[pearl pond]

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[regal herald]

how did 1/2 become 1/4?

[woven kindle]

good point

[regal herald]

what now?

[woven kindle]

is it just square both sides

[regal herald]

that would work, sure

[woven kindle]

so x=16

.close

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[upbeat spoke]

what did I do wrong in this proof?

[brisk scarab]

when you divide by a negative number, you need to swap the direction of the inequality

[upbeat spoke]

thanks

as I get older I realize I miss trivial stuff 😭

If you don't have any other criticisms in the proof, I'll close the channel

[brisk scarab]

Are you trying to prove the top or Cauchy Schwarz?

[upbeat spoke]

the top

I once proved cauchy shwarz couple weeks ago

[brisk scarab]

In that case the logic needs to flow the other way too

[upbeat spoke]

my friend was proud of me

[brisk scarab]

unless those are $\iff$ on the left

[jolly parrot]

Edward II

[brisk scarab]

in which case it's fine

they look more like $\implies$

[jolly parrot]

Edward II

[upbeat spoke]

I don't understand what you're trying to say

I wanted to prove the top inequality true

And after shuffling a bit, I arrived at Cauchy Schwartz

Which I know is true

therefore, the top must be true

[brisk scarab]

did you use $\implies$ or $\iff$ arrows

[upbeat spoke]

isn't that a good enough logic

[jolly parrot]

Edward II

[upbeat spoke]

on the left column? implies

is that wrong?

[brisk scarab]

in that case you're not actually saying that the logic can flow the other way

you're saying if the top is true, then the next line is true, etc, and then Cauchy Schwarz is true

not the other way

like I get what you mean obviously

I'm just pointing out that technically, reading exactly what you've written, you've gone the other way in what you're trying to prove

[upbeat spoke]

I mean, tbh, I never used iff arrows. Even the implies arrow I learnt that from my dad when doing algebra he said to use those to show direction of steps. This isn't a formal proof course. It's a calculus course. The examples use '=' though. is that fine?

but thanks for advice

I'll use iff arrows

It's kinda an old book I should point out

But I'll use modern conventions if it's more accurate

[brisk scarab]

it is, because although here it didn't matter you're not always going to do operations that go both ways

[upbeat spoke]

.close

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[upbeat spoke]

.reopen

[pearl pond]

✅

[upbeat spoke]

wait, so you mean when proving, I should start with something true then get to the thing I'm trying to prove, right?

[brisk scarab]

yes

[upbeat spoke]

🫡

[brisk scarab]

if you start with something wrong

you can always get to something true

[upbeat spoke]

I see

[brisk scarab]

again, the shuffling you did wouldn't have allowed you to change the truth as you go

[upbeat spoke]

well,how though? Unless I'm making mistakes in algebra

oh

[upbeat spoke]

alr

[brisk scarab]

-2 = 2 does indeed imply 4 = 4 by squaring both sides for example

the other way is very much not the case

(in your case you could square without issues to get to line 2 because both sides were positive)

[upbeat spoke]

okie thanks

.close

[pearl pond]

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[chilly slate]

hi I need some help finding the asymptotic equivalent function for
$$\frac{ln(x)cos(x-1)}{\sqrt{(sinh(x-1)}}$$

[jolly parrot]

Mephisto

[chilly slate]

as x approaches 1

from the positive side

I got sinh(x-1) ~ x-1

and cos(x-1) ~ 1

but I have no clue what to do with ln(x)

[burnt dust]

You could try a taylor expansion centered at x=1

[chilly slate]

I think it's x-1?

my letting ln(x) = ln(x-(x-1))

we can expand around 0 if I'm not mistaken

and the first therm is x-1

so we get (x-1)/sqrt(x-1)

when approaching 1 from the positive side, we should get infinity right?

[burnt dust]

The taylor polynomial is in (x-1) when you expand around x=1

[burnt dust]

Not from the expression you got

It is approaching 0

[chilly slate]

ohh right, because our number will become less then one at some point

meaning the root is actually larger

[burnt dust]

This is just sqrt(x-1)

[chilly slate]

ah, true

hm I graphed it

and got this curve

[burnt dust]

Looks good to me

Graph srqt(x-1) on top

Should line up near x=1

[chilly slate]

yup

thanks!

[burnt dust]

You're welcome!

[pearl pond]

@chilly slate Has your question been resolved?

[pearl pond]

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[chilly slate]

hi, I got another question, I have to find a value for a > 0 so that this integral converges
$$\int_1^{\infty}\frac{ln(x)}{x+a} dx$$

[jolly parrot]

Mephisto

[chilly slate]

I'm not sure what to do at infinity?

I get the idea behind this, I just have to find a value for a such that both limits of ln(x) and x+a are equal to each other or a constant multiple

[thorny pulsar]

Are you sure there's an a such that this integral converges?

[chilly slate]

well... I'm not sure

this integral can converge for a < 0 for example

in case there wouldnt be an a > 0, it would diverge

[thorny pulsar]

for a<0 one can apply the comparison test, and since ln(x)/x < ln(x)/(x+a), it diverges

since int (ln(x)/x) diverges*

[chilly slate]

ohhh alright

wait

isnt ln(x)/x > ln(x)/x+a ?

[thorny pulsar]

right

[chilly slate]

wouldn't this mean that you can't use the comparison test?

[thorny pulsar]

sorry give me a second i need to write this out

[chilly slate]

alright

[thorny pulsar]

no, for a<0 ln(x)/x < ln(x)/x+a (at least for x>a, for x<a we can split the integral and do the comparison test for the part that is x>a, since the part that x<a is finite)

[chilly slate]

ah okay

[thorny pulsar]

I think (but am not sure) for a>0 it diverges too, im thinking about this part now

[chilly slate]

so considering we're at infinity

we can disregard the a

no matter what it is

so we end up with ln(x)/x like u said

which converges to 0 is I'm not mistaken

[thorny pulsar]

this one diverges

compare it with 1/x

[chilly slate]

1/x converges to 0 at infinity

[thorny pulsar]

the integral of 1/x diverges

[chilly slate]

ahh like that okay

[thorny pulsar]

this is essentially why i feel it should diverge too

[chilly slate]

alright I think I get it

[thorny pulsar]

anyways, i googled it and you can compute the integral of ln(x)/(x+a) by integration by parts

so after finding the antiderivative you can just check the limit and check if it goes to infinity.

||https://www.mathway.com/Calculus?asciimath=\int ln(x)%2F(x%2Ba)||

[chilly slate]

alright thank you

so to summarise it

ln(x)/x+a ~ ln(x)/x , because ln(x) is big o of x ln(x)/x becomes 1/x of which we take the integral, so ln(x), and ln(x) at infnity diverges

[thorny pulsar]

more or less, but i'd still compute the integral to make sure.

[chilly slate]

alright, thanks for your help

.close

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[pearl pond]

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[lofty stone]

Is there a way to solve this analytically?

[quiet tendon]

you're solving for a?

[lofty stone]

Yes

[quiet tendon]

i don't believe so, if there wasn't an a being multiplied out front then you could turn this into a quadratic

[lofty stone]

x=25 is a given

[quiet tendon]

but with the a out front it's a bit harder

[lofty stone]

This was part of a Calc 2 problem for finding height of some wire post

At the end we all used some sort of graphing software to approximate the answer

But my compulsive self needed an exact solution 🤣

[warm current]

You need the W lambert function

[lofty stone]

If it’s not possible, why is it so? Can we prove it? <@&286206848099549185>

[warm current]

which is not an elementary function

ah nvm, not even W lambert solves it

RIP

[lofty stone]

I see. Thanks for the info. I’ll check that out and see if I can come up with something

[lofty stone]

F

[lofty stone]

Where do you encounter that function?

[warm current]

$x=W(a)$ is defined as the solution to $a=xe^x$

[jolly parrot]

SWR

[lofty stone]

Thanks

.close

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[pearl pond]

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[balmy pilot]

ive currently figured out that mod 7 work

but how do u know which type of mod work in a systematic approach

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[last moth]

Please don't spam the Helpers ping

[balmy pilot]

sorry but can someone pls help

[pearl pond]

@balmy pilot Has your question been resolved?

[pearl pond]

@balmy pilot Has your question been resolved?

[pearl pond]

@balmy pilot Has your question been resolved?

[pearl pond]

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[manic turtle]

sorry r wrong thing

[pearl pond]

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[brittle harbor]

Hi, I have to write a situation/problem (kind of a real world situation but it doesn't have to be very realistic) where we apply the mean value theorem, rolle's theorem and find a maximum or minimum of a function value using derivatives . Can anyone give me some ideas?

[brittle harbor]

I was thinking of someone opening a business (selling and delivering ice cream in a car), and they need to use the least amount of material they can to reduce costs (that would involve finding a minimum value). One day, when driving they exceed the limit speed and have to pay to a police officer (the mean value theorem comes into place here to explain how they knew the person exceeded the speed limit). Only the rolle's theorem would be missing.

[pearl pond]

@brittle harbor Has your question been resolved?

[pearl pond]

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[oblique coyote]

i need help with this

not sure how to solve it

[quiet tendon]

do you know the formula for volume of a cylinder?

[oblique coyote]

yes

I think i found the answer

.close

[pearl pond]

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[pearl pond]

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[minor quest]

Calculus - What is the opposite of a real root.. I was doing the equation of

[minor quest]

Though I couldnt figure out the opposite of a real root as Ive been doing this on my own to teach myself

But Online there is no simple solution/answer to “opposite of a real root”

[tulip ore]

the question is cut off

[minor quest]

Ive solved the question already but in simple terms I was needing to know what the opposite of a real root is

[tulip ore]

there is no such thing as the "opposite of a real root"

you made these words up, they are not standard notation

thats why I need the question so that I know what youre talking about

please post the full question with no cut-off

[quiet tendon]

where is the terminology "real root" even coming in the original question?

unless you're trying to talk about complex roots?

[minor quest]

Im confusing myself

My friend was also mentioning non real roots so thats why I was curious about it because Ive never needed to calculate a “non real” root

[minor quest]

this was the full question it was just a snippet I found and decided to try and do

[quiet tendon]

okay then yeah your friend's referring to a thing called a complex root

if you haven't learned it in school yet then it's probably not worth worrying about, but sometimes an equation that doesn't have a solution in the real numbers has a solution in the complex numbers

like x^2 = -1

has no real solutions but has two complex solutions

[minor quest]

Im in my second year of uni though I didn’t take Spec maths in my senior year so I didn’t learn it - though, im trying to learn it now for the “fun” of it.

[frosty frigate]

if u really want to involve complex numbers, according to the given statement x = 1 is a root but u can say i^4n for n belongs to N is the root

actually n can be any integer

not natural

only

[tulip ore]

do you know what complex numbers are

[quiet tendon]

look up youtube videos about complex numbers and complex zeroes of polynomials

[minor quest]

a+bi like imaginary numbers right

[minor quest]

^

[tulip ore]

bi would be imaginary numbers

[minor quest]

i really only started learning last month so im still getting the hang of it all

[tulip ore]

adding the a+ makes it a complex number

[minor quest]

thank you sm 🙏

[pearl pond]

@minor quest Has your question been resolved?

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[daring wedge]

Which method is more accurate for solving definite integrals
The normal conventional method (yk what I mean)

Or the Newton Raphson method

Or the bisection method
Or the trapezoidal or Simpson rule

[daring wedge]

Which answer is more accurate, the conventional one or the numerical methods one

[regal herald]

numerical methods are often approximations

[quiet tendon]

is conventional using riemann sums? the antiderivative?

[daring wedge]

Yes

[daring wedge]

Yea

Like you normally do

Definite integrals

[regal herald]

the conventional method gives the definite result, like tending the numerical methods to trapezoids of infinitessimal length

for example

[quiet tendon]

well if you can find the antiderivative and solve it conventionally, you have 100% accuracy

[daring wedge]

And numerical methods are just approximations?

[regal herald]

yeah

[daring wedge]

Like they're less accurate than the anti derivative onee

,w integrate e^(x^2)

[jolly parrot]

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Requested by ilovecocomelon_

[regal herald]

there will always be a degree of error if youre using finite terms in them

[daring wedge]

I was told that normal methods don't apply to this integral

Thats why numerical methods are used

[quiet tendon]

your original question very much depends on what the integral is that you're talking about. if it's a function with an easy antiderivative like sinx or x^2 then yeah obviously the conventional approach will have 100% accuracy 100% of the time. if it's something like e^(x^2) which doesn't have an elementary antiderivative then yeah it's gonna be numerically approximated

[daring wedge]

What does the erfi mean