help-13

[south tundra]

Well actually you could use it again

Let me show you

[late spoke]

alright, please

[south tundra]

So the 2nd sum is (n + 1) + (n + 2) + (n + 3) + ... + 49, right?

[late spoke]

aye

[south tundra]

You could rewrite this as (n + 1) + (n + 1 + 1) + (n + 1 + 2) + ... + (n + 1 - n + 48)

But this is the same as 0 + 1 + 2 + ... + (48 - n) + n*(n + 1)

What I did here was take out n+1 from every term of the sum

[late spoke]

i see...

[south tundra]

Oh wait I think it should be (48 - n)(n + 1) there

Because in the sum there are n - 48 terms

Wait let me think for a moment

[late spoke]

ok

[south tundra]

Yeah so the sum is now 1 + 2 + ... + (48 - n) + (48 - n)(n + 1)

We could use the formula here to rewrite this as (48 - n)(49 - n)/2 + (48 - n)(n + 1) I guess

[late spoke]

i got something like thisd

[south tundra]

Is that product of (48 - n), (n - 1), (48 - n) and n divided by 2?

Btw I messed up a bit there, should be (49 - n)(n + 1)

[late spoke]

idk...

this hw is stressing me out

[south tundra]

Yeah so the 2nd sum is basically this

Now all we need to do is set 1st and 2nd sums equal to each other

So what we have is this

The rest is solving for n

[late spoke]

i see

ok, ill see what u can get from here

also sorry to ask u for this, but can u stay here please? i got 2 other questions that im stuck on as well

[south tundra]

Surely

[late spoke]

thank u a ton

[south tundra]

Might get a bit busy soon but I'll try to respond anyways

[late spoke]

its part 2

i asked others in the other hw help qu

but they didnt understand

first of, this isnt an infinite series

its finite to the term of n

second, x is not less that one in this case

[jaunty mural]

differentiation comes to my mind

[late spoke]

cant do diff, have to use xSn

hold on, lemme get the paper out, its like the 3rd time i tried to redo this question

[jaunty mural]

it just seems like it to me but idk

[late spoke]

something like this

but i keep screwing it up somewhere

nvm

[jaunty mural]

wut

1 +x +x^2 + x^3 +...

this is geometric

idk how u got your last line

[late spoke]

idfk either

[jaunty mural]

look this thing is geometric series

write formula

everything before last line is ok

oops

its geometric plus an extra last term

$$1 +x +x^2 + x^3 +... + x^{n-1} -nx^n$$

[wraith dagger]

Shuri2060

[jaunty mural]

I believe, but check.

[late spoke]

got the same as well

...ignore what i wrote on top

ok... i think i got a solution

imma check, by plugging in values x=2 and n=4, i should get 49

i got it....

I GOT IT

AAAAAHHAHAHHAHA

FINALLY

@jaunty mural @south tundra THANK U, NOW SEQUENCES CAN GO KISS MY ASS

HW FINISHEDD

AAAAAAAHHAHAA

THANK U AGAIN

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[cedar kiln]

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[crimson sedge]

hi, i need help

[crimson sedge]

sorry for the bad quality pic

<@&286206848099549185>

<@&286206848099549185>

You would find the height by letting t = 0

height of the bridge?

Yes, because the rock falls from the bridge. So at the beginning, the height of the rock is the height of the bridge

oh okay

How would you find the maximum height of the rock? You could try sketching the quadratic equation

wait so the answer for a is 15 right?

then how is the rock at 25?

Set h = 25 and solve for t

Is it possible? Can you get any answer?

nope

didnt work

Yes, and what did you find as the maximum height?

So for d) you could write 'never'

is that a thing

i think we are doing something wrong

<@&286206848099549185>

[crimson sedge]

Why can't it be a thing? Try drawing the quadratic for the height

I finally understood the whole thing. Thank you!

@cedar kiln

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[deft spruce]

Is the point of inflection on a cubic graph also the midpoint between the minimum and maximum points of the curve?

[cedar kiln]

@deft spruce Has your question been resolved?

[jaunty mural]

Not at all.

Take $x^3+x$

[wraith dagger]

Shuri2060

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@deft spruce Has your question been resolved?

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[humble ledge]

Hi, I'm a bit stuck on trying to figure out if a certain sets of datapoints are generated by a discrete uniform random number generator. The cases I'm having the most trouble with is cases where there are a low amount of samples, or a high amount of possible output values. Ideally I would like to calculate a p value to see if I should reject the null hypothesis that i'm dealing with a discrete uniform generator. I did realize that the amount of times a certain value is hit is distributed according to a binomial distribution, however I am a bit unclear how I should use this to test my null hypothesis.

[humble ledge]

An example of a data set (at least the counts of the elements)
62 48 59 70

(searching for tests that confirm or reject uniformness all lead me to pages about the uniformity of p values)

[steady imp]

What’s your alternative hypothesis

[humble ledge]

so for a binomial distribution with n = 239 and p=1/4 the chance that 48 or smaller comes out is 4.4%

that it isn't uniform?

I don't know if I can make a better alternative hypothesis

[steady imp]

It’s more like

How do you even test that

So like you could fit it?

[humble ledge]

well, a alternative is that it is correlated to other attributes of the datapoint

like if I only look at the data points that also have a certain attribute, I get the counts 11 10 13 26, which my guts say is not uniform, but I can't really prove it

I have one other idea

[steady imp]

I'm looking it up right now

trying to find something mathematical

[humble ledge]

one way of defining a p value would be:
if I generate data sets according to my null hypothesis, what is the probability that my given sample lies further away from 1/4,1/4,1/4,1/4 then this random sample according to a given distance metric

I think that should use the taxicab metric

as then equally likely situations lie equally far (I think)

[steady imp]

So yeah it would be like

something like "there's a p % probability for this particular datum to be situated where it is"

But I'm trying to find a real method fo rit

that for every data point

https://www.investopedia.com/terms/g/goodness-of-fit.asp @humble ledge

https://en.wikipedia.org/wiki/Goodness_of_fit#Pearson's_chi-square_test

This has it fully laid out

So you're basically making a histogram like this

You're histogramming your data into "bins" and comparing the size of each bin to the expected result

and then you should get a statistical answer

Does this make sense?

I think making your number of samples the same number of data points might be good

[humble ledge]

euh

[steady imp]

In that case given your data, your value should be $\chi ^2 = \sum_{i=0}^n{(O_i-1)^2}$ where $O_i$ is the number of data points within the region

Er, that's wrong

[wraith dagger]

KurtDee

[humble ledge]

This is giving me a p value of about 0.98% for 11 10 13 26

[steady imp]

What's your max

[humble ledge]

what do you mean max?

[steady imp]

Like you're sampling uniformly, but like from what uniform distirbution

0-100?

[humble ledge]

0-3

[steady imp]

11 is greater than 3

[humble ledge]

yeah, those are the counts

[steady imp]

OH

oh this is easy then

You already have a histogram lmao

[humble ledge]

right, so chi squared would be the euclidean distance (squared) to [1/4, 1/4, 1/4, 1/4] * n

[steady imp]

I'm not sure I understand

[humble ledge]

to be honest, I think using taxicab distance is more correct

[humble ledge]

euh, nvm, I'm dumb, I think

[steady imp]

I'm not sure what this distance is

[humble ledge]

thanks

[steady imp]

that seems right to me

I hope you understand the math

[humble ledge]

Yeah

[steady imp]

I was just doing it in excel lmao

[humble ledge]

I think the calculator I used used two tail

[steady imp]

Two tailed makes sense

[cedar kiln]

@humble ledge Has your question been resolved?

[deft spruce]

But this graph has no maximum nor minimum points

[cedar kiln]

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[humble ledge]

To answer your question, the inflection is in the middle of the min and max, if they exist

[deft spruce]

Alright, thanks. I think my question clearly implied that the min and max had to exist in order for the midpoint to be in between the min and max points, so I'm not sure why the other person answering me dismissed the obvious implication

[humble ledge]

The inflection point is the max/min of the quadratic differential which is in the middle of the 2 zero points, which are the min and max

@deft spruce also interesting: a cubic function is point symmetric around the inflection point

Wait is it

I'm not sure anymore

[deft spruce]

Wouldn't that imply that the concave and convex part of the graph were identical in size?

OK, and apparently that is consistently true across graphs?
"The graph of a cubic function is symmetric with respect to its inflection point"

[humble ledge]

Yeah, seems like I remembered it correctly

[deft spruce]

Graphs are cool

[cedar kiln]

@humble ledge Has your question been resolved?

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[solemn heron]

do you know what the argument of a complex number is_

[solemn heron]

?

i ask bc i speak another language and i want to make sure its the same term

[drifting matrix]

the intuitiv definition is that it is the angle between the positive real axis and the number.

[solemn heron]

oh yeah its that one

[drifting matrix]

i.e. arg(i) = pi/2, arg(1) = 0, arg(-i) = 3pi/2

[jaunty mural]

The anticlockwise angle from the positive real axis

The range it can be depends on definition

-pi to pi or
0 to 2pi

[solemn heron]

-1 +√3 i ..... i need to get it from this complex number

but when i do the math it gives me -60°

and it should be 120°

the argument

[drifting matrix]

yeah that is different

even if you accout for what Shuri said.

Can you make a sketch?

[solemn heron]

with the cuadrants?

[drifting matrix]

yes

[solemn heron]

i have it

[drifting matrix]

If you sketch it should be clear that 120° is about right

[jaunty mural]

If you're working with complex numbers, you really should use radians not degrees

just going to say

You can think in degrees, that's fine.

[solemn heron]

yeah im using radiants but im not very good with them. its easier to ask in degrees

[jaunty mural]

Anyways

[solemn heron]

but yeah thanks

[jaunty mural]

For the math i dont see how u get that answer

u should sketch on argand diagram

[solemn heron]

i pass it to radiants afte

[drifting matrix]

yes

so more than 90° aka pi/2 and less than 180 aka pi

Actually I think it is always helpful to sketch this or imagine it how it looks

that is how I figure out the right range anyway.

[solemn heron]

and how do i calculate it?

[jaunty mural]

You should work out x

between 0 and 90

[solemn heron]

lets say if i cant see it in the sketch

[jaunty mural]

then subtract from 180

???

Of course you have to sketch

Not sketching is bad

Sketch in head

or sketch on paper, doesn't matter.

I would never do it without.

In particular, you need to know which quadrant the number is in, otherwise you will fail to work out the angle.

[drifting matrix]

You could memorize the cases table but it's not a good idea imho.

[solemn heron]

its the first and forth quadrant is the same

and the second and third are the same

[drifting matrix]

the same?

how do you mean that?

[jaunty mural]

its not.

[solemn heron]

i mean in the first quadrant both imaginary and real are positivw

and the forth both are negative

[drifting matrix]

nooo

[jaunty mural]

I don't think we agree on the quadrant labelling

[solemn heron]

so the process of getting the argument is the same

[jaunty mural]

Standard is to count anticlockwise

Just like your angle.

[solemn heron]

yeah yeah, thats how i see it

[jaunty mural]

The process isn't the same

or if it is

I strongly suggest you don't consider it to be

Just get used to quickly sketching an argand diagram

This takes like 5 secs on paper

[solemn heron]

so its only like a visual thing to calculate it?

[jaunty mural]

yes.

[solemn heron]

oh okay

[jaunty mural]

I can see the angle between 0 and 90 i need to calculate

and then I see if I add? subtract?

to 0, 90, 180, 270, 360

whatever it is

[solemn heron]

yes, i see what you mean

[drifting matrix]

One thing that is maybe worth memorizing is that if it is only at the right side (quadrants 1 and 4), then you can calculate it as atan(imaginary / real)

keep in mind though that it needs adjustment when it is not at the left side.

[solemn heron]

yeah thats how i was doing it

[drifting matrix]

aha

[solemn heron]

but when i do it like that it doesnt give me the 120°

[drifting matrix]

yeah because you are at the left side

as you can see

imaginary / real

gives the same answer

if you negate both.

[solemn heron]

and to do that i just do √3/-1

[drifting matrix]

If it is at the right side you would need to add or subtract 180° aka pi (both are fine depending what definition you prefer)

I suggest to follow Shuri's advice though to sketch also.

[solemn heron]

yeah the thing is i need both

the sketch and the process

and i cant finish understanding the process

i know i should do arctg (√3/-1)

[drifting matrix]

https://wikimedia.org/api/rest_v1/media/math/render/svg/c9bc0c728164372a29cc2801ed3343ac791bcb40

[solemn heron]

but im not sure how to solve it

[drifting matrix]

Unfortunate it does not embed the image

it shows the table of cases if you wanted to figure it out without any sketching at all.

[solemn heron]

oh yeah thanks a lot

🙂

thank you very much

[drifting matrix]

you are welcome

I think this is hard to memorize though without at least sketching in your head.

[solemn heron]

no yeah i know, i always do the sketch first

but if i have to do an exam of this i need to write how i got it

thats why i cant just do the sketch

[drifting matrix]

Hm if I was the teacher I would accept that obviously

but if it does not work for you, oh well

[solemn heron]

hopefully they dont bother much about it

but you are never sure at uni haha

[drifting matrix]

btw I think neither me nor Shuri meant ONLY sketch

we were talking about sketching as a help to figure out the right case.

[solemn heron]

yeah i always sketch

and its rather easy to see how much is the argument

its just that i dont get the math of it

when i do it myself

and i clearly dont know how to use the calculator bc im not getting anything close to that °

120°

but thanks a lot, its really nice that you guys are helping peoplle

ill close it now, have a great day 🙂

.close

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[turbid needle]

I need help determining linear independence, I understand that its making a linear combination of the vectors equaling the zero vector but i'm not sure y i keep getting it wrong.

[turbid needle]

rn i just think A, C, and F are Linear independent <@&286206848099549185>

[brave aspen]

You need to stick the vectors into a matrix equation set equal to the other vector(s)***

And solve

If there is only one solution the vectors are linearly independent

Other pointers

Any set with zero vector is NEVER linearly independent

Even just the Zero vector by itself

So A is not Lin Ind

Also a set that has more vectors than entries in the vector cannot be linearly independent

E.g. set of 3 vectors with only 2 values in them

[turbid needle]

F would also not be LI then

[brave aspen]

Correct

So if you have only 2 vectors

You just try to find a scalar constant that multiplies one into the other

If you cannot find such a number, then those are independent

E.g.

(1,2) ; (3,3)

There is no constant that can turn one into the other

I can solve it algebraically out by entries to be sure but i think you should see that one intuitively

But take a look at

(1,2) , (2,4)

Multiplying the first one by scalar value of 2 gives (2,4)

These are not independent

[turbid needle]

right, is this the only thing im looking for in a set of vectors when determining if its LI?

[brave aspen]

Well if its only 2 then yes

But if you have more than 2 it becomes more complex

[turbid needle]

oh

[brave aspen]

Because e.g.

(1,2), (2,1), (3,3)

Any 2 of those are independent wrt each other

But if you take

(1,2) + (2,1)

You create (3,3)

Adding different multiples of vectors is what we call a linear combination pf those vectors

The real definition of linear independence relies on linear combinations of all the vectors

So if i can add different multiples of vectors together to create one of the others in the set

Then you do not have independence

In order to check for sonething like D.

You need to use a matrix equation

First determine if any 2 of the vectors are independent

[turbid needle]

my work on D

[brave aspen]

This is how i would approach it

Well

Actually your way is fine

For 3 vectors with 3 entries

When you have as many vectors as you do entries, you use the homogenous equation you setup

That equation has only one solution if and only if the vectors were linearly independent

But personally i would stick the first 2 vectors in a matrix and solve for the 3rd vector

If there is no solution, then you have linear independence

I prefer that way of thinking because it works for any size set of any type of vector

[turbid needle]

ok so is C LI since its one vector

[brave aspen]

Yes

G. Is another good example

Definitely NOT LI

Because more vectors than entries in the vector

(Which means more vectors than there are dimensions in that vector space)

[turbid needle]

because the space is Real^2 and there are 3 vectors?

ok i think i understand the content

ty i will answer these now

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[glossy root]

hi so

I have to use gauss elimination to calculate the inverse matrix

and at the end I get the correct answer but with 5s instead of 1s on the left side?

is this still correct?

any help would be appreciated

[drifting matrix]

,wolf {{-1,7},{-2,9}}^-1

[wraith dagger]

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[glossy root]

ohh

I have to divide with 5

bro

[drifting matrix]

Yeah it seems the calculation is correct

until the very last result

[glossy root]

yeh

mb

[cedar kiln]

@glossy root Has your question been resolved?

[cedar kiln]

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[gentle ore]

i solved this question but idk where i went wrong

[tight herald]

"I solved.. idk where I went wrong"

??

[gentle ore]

i "solved" but got wrong answer

so ig i didnt acc solve

lol

[tight herald]

Oh..

[dense wing]

so post your attempt...

[gentle ore]

idfk what i’m doing lmao

[dense wing]

you did all the hard stuff fine

you have a slope (1/6)

and a point (8,3)

[gentle ore]

o i just did line eqn wrong

[dense wing]

yep

[gentle ore]

alr ty lol

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[fair narwhal]

guys does anyone know how to begin proving this is true for all xeR?

[drifting matrix]

What tools do you have available?

[fair narwhal]

what do you mean by tools?

[dawn jetty]

perhaps sin(a+b) identity would be a good start

[fair narwhal]

i tried that but it seems useless

[drifting matrix]

One tool that is incredibly powerful with this kind of question is to use the complex number exp identity but I am not sure if you are allowed to use that.

[jaunty mural]

we wouldnt be suggesting it if it was

[fair narwhal]

you mean sin(x + 2x)?

[fair narwhal]

i think everything is valid

[drifting matrix]

sin(x) = 1/(2i) (exp(ix) - exp(-ix))

[dawn jetty]

going from there to double angle identities maybe ?

it's the most intuitive thing to get sin3x in terms of x

[drifting matrix]

Using this single identity plus usual exponentail stuff I am pretty sure one can do it easily.

But I don't wantto confuse you with complex numbers if you haven't seen them.

[fair narwhal]

no its all good ive heard about them

[jaunty mural]

Just basic compound angle identity is enough.

+common trig identities

[fair narwhal]

thanks for the ideas guys

im gonna try again

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[fair narwhal]

guys how can i know if this series converges?

[west dome]

what tests do you know

for convergence

actually, can you post the full problem first

[fair narwhal]

i know

the leibniz criterion?

[west dome]

this is the test you want to use, yeah

[fair narwhal]

quotient criterion

[west dome]

but can you post the entire question first?

if n is finite then this is trivial

[fair narwhal]

umm

its in german though

[west dome]

thats fine

[west dome]

Okay, so its actually asking you if the sequence (x_n) converges, not if that series in particular converges

[fair narwhal]

ive tried the ratio test but it seems to be 1

so if that inside the sum converges?

[west dome]

this is equivalent to asking if the infinite series converges. It is interesting that they didnt just ask you that outright

[fair narwhal]

so with the sum?

[west dome]

Equivalently, the question is asking if

$\sum_{k=1}^\infty (-1)^k \frac{1}{\sqrt{k}}$

[wraith dagger]

Migillope

[west dome]

converges

Now you can use the Leibnitz criterion (aka alternating series test)

[fair narwhal]

so this means sum from n->inf ?

[drifting matrix]

If there is a (-1)^n thing in it your alarm bells for the Leibniz criterium should always start ringing.

[west dome]

No, thats the sequence of each of those sums with upper bound n

so the first term is $\sum_{k=1}^1 (-1)^k \frac{1}{\sqrt{k}}$

[wraith dagger]

Migillope

[west dome]

the second is $\sum_{k=1}^2 (-1)^k \frac{1}{\sqrt{k}}$

[wraith dagger]

Migillope

[west dome]

etc

so as n gets really really big, itll get really close to $\sum_{k=1}^\infty (-1)^k \frac{1}{\sqrt{k}}$

[wraith dagger]

Migillope

[fair narwhal]

okay i see now

[west dome]

which is why if we can show $\sum_{k=1}^\infty (-1)^k \frac{1}{\sqrt{k}}$ converges, we're done

[wraith dagger]

Migillope

[fair narwhal]

its so difficult because there are so many criteria

[west dome]

yeah, but as @drifting matrix said, if you see a (-1)^k in the series, you should immediately use leibnitz criterion

(I prefer the name alternating series test because it makes it a lot easier to remember what it actually does)

[fair narwhal]

i see

[west dome]

anyway, do you see how to solve this using the leibnitz criterion?

[fair narwhal]

so this would be alternating too?

[west dome]

no, there is no (-1)^k here

[fair narwhal]

oh i thought because of cos()

okay but youre right I'll have to try to solve a) first

[west dome]

no matter what power you raise e to, it will never be negative

so there is no alternating ness

although, you might want to examine part b by saying "well if the exponent converges, then the sequence converges"

then you can probably use alternating series test

but we can cross that bridge when we get done with a)

[fair narwhal]

ok thank you very much

[west dome]

ping me when you are done (or stuck at some point) on a

[fair narwhal]

.close

[cedar kiln]

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[fair narwhal]

@west dome hey i think I did it:
first: i set 1/sqrt(k) as my x_n
second: i did the lim of x_n, which was 0.
third: i proved if x_n >= x_n+1, which was true

[west dome]

Looks like the right process!

You ready for b)?

[fair narwhal]

but im not sure how that proved a)

[west dome]

oop

[fair narwhal]

is it possible to find the actual value it converges to, or would that be too difficult?

[west dome]

you could. my german is rusty, but the question just asks if it does converge, right?

[fair narwhal]

yep

Okay i guess im just gonna trust these 3 steps

[drifting matrix]

I think finding the limit would be difficult in this case.

,wolf sum (-1)^n/sqrt(n), 0<n<oo

[wraith dagger]

Results provided by WolframAlpha

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

Requested by M8732#6402

[drifting matrix]

basically wolfram alpha is cheating

gamma function is just defined like this.

[west dome]

They're trustworthy steps, seeing as how it is the alternating series test, which isnt too hard to prove!

[fair narwhal]

Alright

[west dome]

so @fair narwhal did you want to do part b?

[fair narwhal]

yes

[west dome]

ok, do you have anny ideas

[fair narwhal]

there is another tip for this i think: if no limit exists, check if an accumulation point exists

[fair narwhal]

so something with this

i can maybe just use the cos pi sqrt(n)?

[west dome]

yeah so we have the condition that exp(cos(...)) converges iff cos(...) converges

so: you need to test if cos(...) converges

[fair narwhal]

ah but cos never does

[west dome]

cosine never does, true, but cos(1/n) surely does, right?

for example

[fair narwhal]

Ok true

to either 1 or -1 right?

[west dome]

to 1

since 1/n -> 0 as n-> infinity and cos(0)=1

so, you need to look closely at pi*sqrt(n)

does this converge?

[fair narwhal]

nope

but i dont understand, which criteria should be used now?

[west dome]

uh... logic. I'm not sure the name of the criterion, I can check if youd like

Divergence test

there it is

[fair narwhal]

Alright

so as you said its e^cos(inf) basically?

[west dome]

well, its a little more complicated

if the inside of the cosine had the sequence 0, 2pi, 4pi, etc, then the inside diverges to inifinity, but cos of the sequence is constant (1) right?

essentially, you need to show that $\pi sqrt{n}$ doesn't converge and that it doesn't start repeating values for which cosine is constant

[wraith dagger]

Migillope

[west dome]

i.e. doesnt look like some $\theta + 2\pi k$

[wraith dagger]

Migillope

[west dome]

well, I suppose it could alternate between values on which cosine has the same output, but it doesn't need to be so complicated. Do you see what I'm getting at?

[fair narwhal]

No

[west dome]

ok, let me draw a pic

[fair narwhal]

because for me it looks like an alternating series

[west dome]

the cosine alternates, the e does not

[fair narwhal]

oh yea true e is always positive

ok i see now

[west dome]

you can probably use alternating on the cosine, but its complicated

anyway, cosine goes around this cirlce

if the argument of cosine (here, $\pi \sqrt{n}) keeps landing here, what happens?

that should be pi/4

[fair narwhal]

i dont understand, how should i know what cosine(pi*sqrt(n)) is

[west dome]

you mean cosine(pi*sqrt(n))?

[fair narwhal]

yea sorry

i just know that cos(...) is either + or -

[west dome]

Okay lets work from there, why is it either + or -

[fair narwhal]

because cosine is either + or -

at least for pi*sqrt(n)

for lets say 2* n *pi its always 1 like you said

[west dome]

right! So the point is that

1. pi*sqrt(n) does not converge
   and
2. It doesnt start repeating values on which cosine is constant. We can show this by nothing that if you take the subsequence where n = 1, 4, 9, 16, etc, our sequence becomes $(\pi \sqrt{1}, \pi \sqrt{4}, ...)$ which is $(\pi, 2 \pi, 3 \pi, ...)$

[wraith dagger]

Migillope

[west dome]

make sense so far?

[fair narwhal]

yep

[west dome]

okay, now what is cosine of that last sequence

i.e., $(\cos(\pi), \cos(2\pi), ...)$

[wraith dagger]

Migillope

[fair narwhal]

-1, 1, -1, 1

[west dome]

right!

since we we have a divergent subsequence, the entire sequence is divergent

this shows that $cos(\pi \sqrt{n})$ is divergent.

[wraith dagger]

Migillope

[west dome]

then, of course, $e^{cos(\pi \sqrt{n})}$ is divergent

by any number of tests

[wraith dagger]

Migillope

[west dome]

@fair narwhal would you like to go over it one more time or do you get it?

[fair narwhal]

no it does make sense

but what are the accumulation points then?

[west dome]

does it ask for that?

[fair narwhal]

sadly

[west dome]

uh, well. The sequence diverges.

[fair narwhal]

wait but cos can only be -1 or 1 as their maximum points

nvm idk

[west dome]

accumulation points are the limit of the sequence

[drifting matrix]

(b) has infinitely many accumulation points, every point in [-1,1] is an accumulaton point.

but this is pretty difficult to proof rigourously I think.

[west dome]

? the sequence diverges, it has no accumulation point

[drifting matrix]

no

accumulation point just means that it touches any neighborhood infinitely often.

As the picture says in german

[west dome]

ah, I was looking at the wrong defintion

[drifting matrix]

"if NO limit point exists, figure out if there is an accumulation point"

[fair narwhal]

can you explain that for dummies

[drifting matrix]

Luckily it does not ask for a specific accumulation point

so there is a simple argument that there exists one without finding one concretely.

[drifting matrix]

what part exactly?

[fair narwhal]

how i can see if theres an accumulation point

we always did it with lim sup, lim inf i think

[drifting matrix]

First is it clear to you what an accumulation point is?

[fair narwhal]

ummm

[drifting matrix]

Maybe we should talk about this first?

[fair narwhal]

a point where many (infinite?) points touch it

[drifting matrix]

yes intuitively

there are at least 2 formal definitions:

• there exists a subsequence which converges to this point

• a point x such that for any epsilon > 0 and arbitrary N, there exists n >= N such that |x_n - x| <= epsilon

[fair narwhal]

the what now????

[drifting matrix]

it's kind of the same thing

[fair narwhal]

yea i stopped trying to figure out the epsilon thing a long time ago

[west dome]

that's not great

[drifting matrix]

:/

indeed

[fair narwhal]

maybe i'll understand it next semester

[drifting matrix]

For our intents and purposes we can work with the first definition, which hides the epsilon inside the convergence.

[fair narwhal]

ok great

[drifting matrix]

so we basically discard any sequence elements we don't like to get a convergent sequence

let me give a simple example

[fair narwhal]

so if the sequence diverges all the subsequences diverge too

[drifting matrix]

x_n = (-1)^n

this is not convergent as you can see (maybe draw a picture, it goes between -1 and 1 all the time infinitely often)

[fair narwhal]

ok right

and the accumulation points are -1 and 1

[drifting matrix]

yep

[fair narwhal]

makes sense

[drifting matrix]

they are touched infinitely often

using subsequences x_2n and x_2n+1

the even and odd sequence entries

okay

So I think for our specific example we just need to know one important property of sequences

which is often called compactness or Bolzano Weierstraß property

it says that if a sequence is bounded then there exists at least 1 accumulation point

The intuitiv idea is that the elements must be "somewhere" and since the sequence is bounded they can't escape

so somewhere there gotta be a lotta elements

[fair narwhal]

thats true

[drifting matrix]

Given this property you get that there is an accumulation point

without doing any real calculation

[fair narwhal]

wait thats so OP

[drifting matrix]

🙂

[fair narwhal]

but the series is divergent

or doesnt that matter?

[drifting matrix]

if it is divergent to infinity then you cannot apply it

since it's not bounded

actually if it divergent to infinity there will be no (real) accumulation point

same with minus ininifty of course

[fair narwhal]

Ok

but in this case it works because

cos is bound to -1 and 1?

[drifting matrix]

yes

and exp increasing

so

1/e = exp(-1) <= exp(cos(...)) <= exp(1) = e

[fair narwhal]

Ok

but those are not the 2 accumulation points

we dont know them right?

[drifting matrix]

yeah as I said I am pretty sure every number between 1/e and e

is actually an accumulation point

[fair narwhal]

that would make sense yea

[drifting matrix]

but luckily we don't need to proof that

[fair narwhal]

alright i understood it now

[drifting matrix]

yay 🙂

[fair narwhal]

thanks to the both of you

[drifting matrix]

You are welcome

[fair narwhal]

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[west dome]

indeed, since our function is continuous and oscillating between these two values, every point between is accum. pt

[west dome]

.close

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[odd wharf]

how do I find the anti-derivative of x / (x^2 + 4)?

[odd wharf]

none of the rules on my trig sheet apply to this

[glad kestrel]

looks like a u sub

[odd wharf]

Symbolab said that

But I don't understand how that works with the problem

[glad kestrel]

you have the derivative of your denominator as the numerator, off by a constant

normally calls for a u sub

[odd wharf]

uh

ok so int [ x / (x^2 + 4) ] dx -> u = x^2 + 4 -> du = 2x dx -> int [x/u] 2x dx

[glad kestrel]

it's a search and replace type thing

$\frac{\dd{u}}{2}=x\dd{x}$, agree?

[wraith dagger]

a disappointing son

[odd wharf]

wouldn't it be du/2x? not just over 2?

[glad kestrel]

well i can just divide by the 2 here

still the same thing, just keeping x on the other side

makes it a bit easier to see

[odd wharf]

wait huh?

oh

x/u * du/2x

x's cancel out

[glad kestrel]

they do, yes

[odd wharf]

1/u * du/2

ok so

[graceful karma]

u sub makes me shudder

[odd wharf]

1/u * du/2 = 1/2 * int[ 1/u du]

[glad kestrel]

looks good i think

[graceful karma]

Calculus class leaves scars

[glad kestrel]

don't diss the u sub :(

[odd wharf]

which is 1/2 * ln |u| according to the anti-deriv sheet

[glad kestrel]

indeed it is

[graceful karma]

Calculus students when the integral isnt set up perfectly for u sub to work

[glad kestrel]

[gaunt hamlet]

Engineers when they're not given a sheet with the answer to every integral in existence:

[graceful karma]

Just tape one to the wall

[cedar kiln]

@odd wharf Has your question been resolved?

[cedar kiln]

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[zenith python]

.close

[cedar kiln]

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[vale jackal]

Hilf mir

[frigid canyon]

kannst du sprechen englisch, bitte

in which question?

[cedar kiln]

@vale jackal Has your question been resolved?

[cedar kiln]

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[dreamy plume]

what are the possible values of k expressed as an inequality if y=-x+3 and y=k(x+1)-1 intersect in the first quadrant

[remote parcel]

well

first lets graph it in desmos

we know

that we can find a range for k

by finding the value of k that would result the second equation passing through (3, 0)

and by finding the value of k that would result the second equation passing through (0, 3)

the rest is just algebra

[dreamy plume]

ok thanks

[cedar kiln]

@dreamy plume Has your question been resolved?

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[warm vector]

How would I do this question?

[livid hound]

properties of the discriminant

[warm vector]

But There’s an extra variable that’s confusing me

[livid hound]

consider b+1 to be the constant component of the quadratic

[remote parcel]

as long as the D is greater than zero or equal to zero

b^2-4ac has to be greater than or equal to zero

now

the b doesn't actually mean the b in the equation

it means the b as in ax^2+bx+c=0

[warm vector]

Ok

I don’t really get it

[livid hound]

which part don't you get

[remote parcel]

1-4(2)(b+1)>0 or =0

[warm vector]

How to start solving it

[livid hound]

properties of the discriminant

do you know what the discriminant of a quadratic is

[warm vector]

B squared -4ac

[livid hound]

do you know what the variables in that expression represent?

[warm vector]

Yes

[livid hound]

can you apply that to the quadratic you're given

[warm vector]

Except in this question I don’t get what’s the variable C representing

[warm vector]

Yes

[livid hound]

do you know what the variables in that expression represent?
yes
Except in this question I don’t get what’s the variable C representing...
that's a bit contradictory

c represents the part independent of x

[remote parcel]

b+1

[livid hound]

which people refer to as the constant

consider b+1 to be the constant component of the quadratic

[warm vector]

Oh ok I get it

Ok I got it thanks

I have another question

How would I do this question?

[livid hound]

same idea

properties of the discriminant

[warm vector]

So what would the value of B be?

1 I think

[livid hound]

yes

[warm vector]

Ok

And the value of A would be k+2 I meant?

[livid hound]

yes

[warm vector]

Ok right

Thanks

For your help

.close

[cedar kiln]

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[cedar kiln]

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[dense dawn]

Hi! I have a question on this problem; I don't have any work available at the moment as I don't understand what the question is asking. Is the first step to set the equations equal to each other and then solve from there? But even then, I wouldn't know what to solve for.
(I'm new to the server, please let me know if I'm doing something wrong)
I'm going to sleep soon so it might take me a while to respond after some time

[cedar kiln]

@dense dawn Has your question been resolved?

[dire geode]

damn teacher why you giving proofs to 8th graders

[dire geode]

start with $ax^2 + bx + c$. do algebra on it until you get $a(x-h)^2 + k$. first step might be to factor out $a(x^2 + (b/a) x) + c$.

[wraith dagger]

riemann

[dense dawn]

So basically turn standard form into vertex form?

[dire geode]

at the end of the algebra, you'll be able to write $h$ as something with $a, b, c$ and same for $k$

[wraith dagger]

riemann

[dire geode]

correct

[dense dawn]

Ohh

[west dome]

8th grade teacher assigned this?? sheesh.

[dense dawn]

Nah, 9th grade lmao

[west dome]

okay still

[dire geode]

oh then that's completely fine /s

[west dome]

my undergrads cant do this

[dire geode]

lol

Zizi's in honors algebra for sure

80% of her high school grads go to Harvard/MIT/Princeton/Stanford

[dense dawn]

PFFF

Okay so I got $a[(x+(b/2a))^2-(b^2/4a^2]+c$

[wraith dagger]

Zizi_Sunee

[dense dawn]

$a$ is the same, $h=b/2a$, and $k=(-b^2/4a)+c$ i think

[wraith dagger]

Zizi_Sunee

[dire geode]

you can always verify by plugging it in and simplifying

that's my way of telling you that you're almost there

[dense dawn]

hmm ok so now i have to find a, b, and c right

[dire geode]

i mean plug in your values of $h$ and $k$ into the vertex form

[wraith dagger]

riemann

[dire geode]

$a \left(x - \frac{b}{2a}\right)^2 + \frac{-b^2}{4a} + c$

[wraith dagger]

riemann

[dire geode]

does that equal the standard form?

[dense dawn]

no? (guessing at this point)

[dire geode]

gotta do the work zizi

you're a future princeton student, gotta work as hard as them

[dense dawn]

ok so thats the values plugged in, right?

[dire geode]

those are your values, yes

[dense dawn]

i don't really understand what i'm supposed to do next, though 🤔

[dire geode]

expand this and see if it equals $ax^2 + bx + c$

[wraith dagger]

riemann

[dire geode]

you can think of it as a "verify" step

[dense dawn]

hmm alright

yes!

[dire geode]

this is equal to $ax^2 - bx +c$

[wraith dagger]

riemann

[dense dawn]

yes

[dire geode]

you're supposed to get this

[dire geode]

you just have a sign error on h here.

[dense dawn]

oh yeah wait gimme a sec

yeah ok i see it now

so $a(x-(b/2a)^2)+(b^2/4a)+c$

[wraith dagger]

Zizi_Sunee

[dense dawn]

whoops i meant $a(x-(b/2a))^2+(b^2/4a)+c$

[wraith dagger]

Zizi_Sunee

[dense dawn]

Oof I gtg sleep

[dire geode]

so close!

it's the other term

$(x+h) = (x - (-h))$ my last hint

[wraith dagger]

riemann

[cedar kiln]

@dense dawn Has your question been resolved?

[cedar kiln]

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[pliant sandal]

.close

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[spiral onyx]

The number of solutions of the equation [2x]-[x+1]=2x (where [ ] is the greatest integer function) is?

[harsh dove]

@spiral onyx You know what greatest integer function is?

[spiral onyx]

yes but im kind of confused about the question

like i haven't solved such questions on GIF before

[harsh dove]

Can you find at least one solution? @spiral onyx

[spiral onyx]

umm

im not sure

[harsh dove]

That would be my first attempt on doing such guestions.

For example 0 is the solution or not?

[spiral onyx]

how is 0 a solution....

[harsh dove]

I'm asking you, is 0 a solution or not?

[spiral onyx]

i don't think so

nvm it's okay I'll ask my teacher tomorrow

.close

[cedar kiln]

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[indigo plaza]

it is in the 10th power

how can i find a pattern to calculate this

[dense wing]

check if it's diagonalizable

[indigo plaza]

oh i see

[cedar kiln]

@indigo plaza Has your question been resolved?

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[haughty gyro]

just a quick question, would i need to use u-substitution to integrate this?

[haughty gyro]

im pretty sure its basic rule but idk

[fallen heath]

nuu, it's standard integration

[azure crater]

hmm, i think it would be easier if you multiply 2 with x^3 and -x and then you just need to integrate each term apart

[iron berry]

you just multiply it out, split the integral and just integrate the polynomials

actually you dont even need to multiply it out since integrals are linear

[haughty gyro]

ty

do i need to split them or is that just to simplify

its in the context of this problem

[graceful karma]

Or just extract the constant outside the integral

[haughty gyro]

ah

Seems like i messed up somewhere

[graceful karma]

Wouldnt they want you to evaluate it

I guess that doesnt evaluate to the right answer

Also whyd you split the integral?

[haughty gyro]

it doesnt care if i evaluate so i usually dont

just so theres less margin for error

and i split it because of the two different areas

[graceful karma]

Id just do it in one integral

Since its all one function

[gaunt hamlet]

You have to deal with that negative area, no?

[haughty gyro]

yeah so i set it up as

well idk how to use the bot to write it so ill take another picture

[graceful karma]

,rotate ccw

[haughty gyro]

srry

its the bottom one

the integration from -1 to 0 + the integration from 0 to 1

[graceful karma]

Why not just fundamental theorem it

F(1)-F(-1)

Rather than F(1)-F(0)+F(0)-F(-1)

[haughty gyro]

ohhh

[graceful karma]

The F(0) cancels out anyways

[haughty gyro]

alright ill try that

i still got it wrong

i got ((1/2) - 1) + ((-1/2) + 1)

[gaunt hamlet]

You have to do the integral from -1 to 0 MINUS the integral from 0 to 1 to account for the negative area, I think

[haughty gyro]

didnt work either

[gaunt hamlet]

What did you input?

[graceful karma]

The integral would already be negative

F(-1) for example would be -0.5

[gaunt hamlet]

But we don't want the signed area, no?

[haughty gyro]

i just tried subtracting instead of adding

[gaunt hamlet]

So we gotta account for that?

[graceful karma]

Oh is that what they want?

[gaunt hamlet]

I'm assuming based on the wording

[graceful karma]

Usually Id think that meant the negative area too

Irdk though

[haughty gyro]

i figured it out

i missed a negative sign somewhere

ty

[graceful karma]

Wait was it total area?

Or like positive negative

Now Im curious

[haughty gyro]

it was total area

so the first part on the left was 1

and the second part was 1 as well

answer was 2

i honestly thought they would cancel out but here we are

[gaunt hamlet]

They would cancel out if it was signed area

[cedar kiln]

@haughty gyro Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[umbral marsh]

how do I solve

[umbral marsh]

Im making a function that returns a and b

x1, y2 can anywhere, behind it, above it, below it, in front of it, etc.

[crimson sedge]

Oh. Uh one sec i think i can help with this one

[umbral marsh]

first I need the angle I'm pretty sure

ok now it looks better

[crimson sedge]

you could go about it using trig

[umbral marsh]

I just need a and b's values

[umbral marsh]

I know

[crimson sedge]

but i think you can also go about it using scaling i think, that might be easier, do you want that way?

[umbral marsh]

just any solution I can understand / implement in programming

[crimson sedge]

do you understand basic trig functions?

[umbral marsh]

yes

sin, cos, tan, asin, acos, atan and the cs I cant memorize their names

[crimson sedge]

oh that does noy have have a back ground

hold on wait, i really think this will be easier with just scaling

[umbral marsh]

yes I know this...

[crimson sedge]

give me a moment to work out a gaood way to explane

so if we have a rectangle and we cut the side lengths in half the resulting red rectangle is 1/4 the size in area

and if we make the length of the rectangle 1/3 then the area is 1/9

what im trying to show here is that there is a relationship between scaling so and the area

[umbral marsh]

k

i think I get what ur saying

what's the scale in this

I can figure out longer sides

and the long line in the middle

I know the small line in the middle in the small rectangle is 10

[crimson sedge]

so if we make the rectangle
1/2 length the area ia 1/4
1/3 length the area is 1/9

so if se scale the rectangle to
1/x length the area is 1/x^2

ON WAIT

the length of that line is 10? no the area of the small rectangle?

[umbral marsh]

yeah

length of that line is 10

sorry

I didn't clarify

[crimson sedge]

lmao, i was assuming the area of that square is 10

this is a much easer question now lol

when you uniformly scale a triangle the ratio between the side length stays the same

[umbral marsh]

ok one sec

[crimson sedge]

let me know when ever you ready to contine

[umbral marsh]

LOOOOOLL

THIS QUESTION IS SO EASY

WHY DID I NOT THINK OF SCALING

TYSM

.close

[cedar kiln]

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[keen thunder]

Is 24° correct?

[keen thunder]

If not what's the correct answer?