help-17

[vast shale]

,rotate

[vast shale]

very hard to read the question :(

what's the original question

[lost barn]

When u get $\int \frac{\dd u}{\sqrt{u - u^2}}$

[twin meteor]

hiidostuff

[lost barn]

Try completing the square on the bottom

Oh nvm u did

[tight shard]

Integral of square root of 1+1/sinx

[vast shale]

just to confirm

$$I = \int \sqrt{1 + \frac{1}{\sin{x}}} dx$$

this right?

[tight shard]

All under the square root

[twin meteor]

Edmund Cloudsley

[vast shale]

correct?

[tight shard]

Right

For clarification, my first step was multiplying the argument under the square root by 1-sinx/1-sinx

So i get ( cosx)^2 and get rid of the square root in denominator

[vast shale]

yup that should work

[tight shard]

The next step is u-sub

Eliminating cosine

Not sure whether subbing 1-sinx was right decision though

[vast shale]

it does look to be correct

since you have to eliminate the cos(x) on the numerator eventually

[tight shard]

So my result is correct?

[vast shale]

Unfortunately, I cannot make out what you have written

[tight shard]

Where did you get lost?

[vast shale]

I can't differentiate the x and the u

could you perhaps re-write this solution on another page

and post it here?

[tight shard]

Can I rewrite only the part after subbing?

[vast shale]

Sure

Sorry I know this would take some more of your time

but it would help me point out the mistake much faster

[tight shard]

Its alright, iam really thankful for your help

[vast shale]

feel free to ping me once you are done

[tight shard]

Sure

@vast shale

Any conclusion? 😆

@vast shale

[vocal sleet]

@tight shard Has your question been resolved?

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@tight shard Has your question been resolved?

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[desert glen]

How did they go from 2^-ln n to 1/n^ln2

[desert glen]

this is as far as i got lol

[grizzled vault]

,, 2^{-x} = \frac{1}{2^x}

[twin meteor]

mmmm7

[desert glen]

Yeah, thats what I did but $\frac{1}{2^(ln n)}$ isnt $\frac{1}{n^(ln 2)}$

[twin meteor]

dingy

[grizzled vault]

wdym?

x -> -ln(n)

[desert glen]

look at the change in exponents

[halcyon ice]

2^-ln n = 1/(2^ln n) =1/(e^ln2)^(ln n) = 1/e^(ln2 * ln n) = 1/(e^ln n)^ln2) = 1/n^ln2

[grizzled vault]

oh wait i'm reading your work

instead of what you're asking

[desert glen]

lol

[grizzled vault]

😭

[mental egret]

2^{-ln n} = 1/(2^{ln n}) = 1/(n^{ln 2})

Are you asking about why the second equality holds?

[desert glen]

im more so asking how they went from that one form to the other

i dont get how this step was made

[halcyon ice]

its a rule of exponents and logarithms

x = e^(lnx)

since the exponential function and the logarithm are inverses they cancel each other out

in this case we go from 2 to e^Ln2

then just slap on the 1/ and ^ln(n)

[desert glen]

I see

Thank you!!

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[vast shale]

Does anyone know where I can learn this?

[vast shale]

It's barely explained in the book, I don't even really know what it's called haha. When I try to google 'Set theory union infinite intervals)

Or something like that, i don't find anything

If anyone can refer me to a good youtube video or something that explains these proofs wel, it'd mean the world

I never understand how they make these proofs

[tidal dock]

i don't know any material on this but i can help you with the actual problem

[vast shale]

That'd be amazing!

Wait, is it cool if I then quickly get a new exercise that's basically the same but with different values?

Since I've already seen the answer of this one haha

[tidal dock]

yes

[tidal dock]

okay so

$\frac{n+1}{2n} = \frac{1}{2} + \frac{1}{2n}$

[twin meteor]

artemetra

[tidal dock]

it's a bit easier to work with

[vast shale]

I know the first step, namely, that this intersection should be equal to (1, 3] indeed

[tidal dock]

yep

[vast shale]

Yeah that makes it way easier indeed

[tidal dock]

so you are struggling with the proof?

[vast shale]

Yes, very

[tidal dock]

okay so i'm gonna go by the same steps as here

[vast shale]

That's cool

So first take an arbitrary a inside the intersection

[vast shale]

By the way, I gotta say, the proofs differ quite a lot for each exercise, so it might backfire

That's why I struggle a lot with them haha 😅

[tidal dock]

okay i'm back

[vast shale]

Hi, I've tried to do it your way, so like just mimic the exercise above

,rotate

[vast shale]

This is half of the proof though

[tidal dock]

right so this establishes that the interval we wanna work with is (1, 3]

(i would honestly accept this as good enough of a proof but ig your school wants you to be more rigorous)

[tidal dock]

this proves the $\subseteq$ part

[twin meteor]

artemetra

[tidal dock]

now for the $\supseteq$

[twin meteor]

artemetra

[vast shale]

Wait, before me continue

[tidal dock]

yes?

[vast shale]

I've just taken a look at the answer for the first part

Would you think it would also be correct what I have?

[tidal dock]

wait reverse those lol

oops

[vast shale]

OOPS

😂

I'm sorry I looked at the wrong part

[tidal dock]

no no you were at the correct part

[tidal dock]

yes

your proof is good for this

[vast shale]

No right? this part takes a inside (1,3]

However I took a inside the intersection

[vast shale]

that was my proof

[tidal dock]

....right

[tidal dock]

indeed

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[tidal dock]

it's literally your proof but backwards haha

[vast shale]

Oh it is indeed haha

[vast shale]

Is it okay though? That I didn't do that proof on the other side

I'm sorry if I'm asking dumb questions I'm a bit confused why they like utilize strategies differently everytime

[tidal dock]

i see that you have the idea but it's not quite what they want from you

[thick ocean]

Thanks I was actually offline that time

[tidal dock]

if it's a question on intersection, then
whenever they ask you for the ">" part, you pick 'a' from the set it you want it to be in and show that for any 'a', it's in the intersection, so all 'a''s in (1,3] are in the intersection
then for the "<" part, you let 'a' be the intersection and then in particular the biggest 'a' can possibly be, which here is (1,4) and then you rule out all the choices that make it differ from (1,3], in particular you figure out that 'a' can't be bigger than 3

[vast shale]

Wooow okay

I'll try to apply that in the practice exercises

How would it work with unions?

I can show you 2 examples of proofs of unions

[tidal dock]

that's the fun part – for unions it's almost the complete opposite!

[vast shale]

I'm just putting them here so it's easier to view haha

Let me see

[tidal dock]

for unions
the ">" part: you pick x to be from the set you want it to be, and then show that for some element index n or k, x will be in the corresponding set with which we take union with.. in other words: for all x, find k such that it's in the interval, which immediately means it's in the union
the "<" part: you pick x to be from the union, and then simply show that it cannot be either bigger than or smaller than the endpoints of your target set. this is sufficient as if bounds all possible x's from the union

but you typically see "<" before ">" because it's easier

[tidal dock]

all correct!

[vast shale]

I dont really understand this though, for the < part, so x in the interval you found, herethey say take x => 1 then ... and x < 1

why is this a smart choice

Cause I would've never foudn this myself

And do I do this for every proof with unions?

[tidal dock]

i'm not sure what you are referring to

the " take x => 1 "

[vast shale]

I'm sorry I eman a => 1 and a < 1

So this:

[tidal dock]

ah

right so that's in the > part

[vast shale]

Oh yea, > indeed haha

[tidal dock]

yeah they are splitting the cases. if a >= 1, then x=1 immediately works and we don't really need to do anything, because that interval is [1, 5)

and now the other case is when a < 1, which means a is in (0,1), which is the same as the parameter x!

so we can simply take x=a because [a, 3x+2) certainly contains a

yeah i agree that was rather smart of them

[vast shale]

Ohh okay

This only works because the union is only x inside (0,1) of course

Final question, what about this?

the > part again

I don't understand the x<1 part again haha

[tidal dock]

your goal is given $x$, get a $k$ such that the inequality $$\frac{1}{k+1} \leq x < 1 + \frac{1}{k+1}$$ is satisfied

[twin meteor]

artemetra

[tidal dock]

does that part make sense?

my real analysis professor taught a very useful approach with these problems/proofs, and that is establishing what the actual goal is, or is equivalent to

[vast shale]

Yes, it makes sense idneed

[tidal dock]

awesome

[vast shale]

its the same as saying x is inside that interval

[tidal dock]

yep

so we also don't really care about the 1+1/(k+1) part either

[vast shale]

My Analysis professor told me we don't teach you proofs, weteach you some handy tools you can use and you become skilled in them yourself as you encounter them again and again haha

[tidal dock]

so we basically need any k s.t. 1/(k+1) <= x

let's rearrange the inequality and solve for k

1 <= x(k+1)

x(k+1) >= 1

k+1 >= 1/x

k >= 1/x - 1

so any such k works, and since k is an integer they round it up to make it an integer

[tidal dock]

damn

[vast shale]

Wooow

[tidal dock]

different styles

[vast shale]

I understand how you did it indeed

[vast shale]

However, why don't we use this

[tidal dock]

because x < 1 and 1 + 1/(k+1) is strictly greater than 1

[vast shale]

OHHHHHHHHHHHHHHHH

[tidal dock]

it's always true

[vast shale]

i love you youre the best

[tidal dock]

haha

you are welcome bro

[vast shale]

thxx man

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[boreal remnant]

so in deriving the product rule (using differentials) we have that:
d(uv) = (u+du)(v+dv)-uv = u dv + v du + du dv

[boreal remnant]

so the question is why do we ignore squared differentials?

[half imp]

Because they do not measure rate of change

[heavy yoke]

if we imagine du and dv as being both very small quantities, then the product of two small quantities will be negligible

[half imp]

They are infinitesimal in comparison to the things we're interested in

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[cosmic mountain]

would question 22 involve repeating the steps in 18-21 but with 785.0kv instead of 500

[cosmic mountain]

im confused because the question seems to go specifically into voltage and power lost to resistence and idk how this ties into personal electricity use which they seem to be refering to

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[bronze osprey]

you need to use the equation $P = I^2 R$ to calculate the power loss

[twin meteor]

south

[bronze osprey]

oh wait you did that in q20

yes then that's correct, calculate the new current and then the new power lost to resistance

then repeat what you did in q21

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[desert hornet]

Once again is there an easier way of doing this than writing out an explicit injective map?

this is 3.15 btw

[hard atlas]

its not like two lines is too much to write

I cant think of another proof right now which doesn't do that

[desert hornet]

It isn't, I just don't like writing out linear maps in particular

Maybe I should just get used to it

[hard atlas]

but its so nice to write out a linear map

only need to pick a basis and send it somewhere

[desert hornet]

that's true

it just feels a bit like cheating I guess

pretty much the same idea

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[hard atlas]

yes

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[distant perch]

I didn't really understand the statement

[distant perch]

Can you please explain me how I should start the solution

I actually think I understood. Joe's ratio is 10/2 because he drank 2 ounces of coffe and added 2 ounces of cream and JoAnn's ratio is 11/1 because she added 2 ounces of cream: 12/2, stirred well and drank 2 ounces. If she stirred well, the 2 ounces that she drank have the same amount of coffe and cream, so she drank 1 ounce of coffe and 1 ounce of cream. So Joe's ratio = 10/2 and JoAnn's ratio = 11/1. I multiplied Joe's ratio by 11 and JoAnn's ratio by 10, so it is 110/22 and 110/10. Joe's cream divided by JoAnn's cream is 22/10 = 11/5

Am I right?

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[nocturne spade]

What is your question?

[desert hornet]

Just checking if my proof is correct and if I can word it better

[hard atlas]

the implications at the start are a bit worrisome

why is Tv=0 equivalent to what you wrote in the line above that

[desert hornet]

S(Tv)=0 if and only if v=0, so Tv=0 is only possible when v=0 as otherwise we would have a non-zero v with S(Tv)=0

[hard atlas]

but is that an equivalence?

which direction of implications are you using here

[desert hornet]

$\big((ST)v=0\iff v=0\big)\implies (Tv=0\implies v=0)$

[twin meteor]

kheerii

[hard atlas]

and how does the left imply the right?

[desert hornet]

Because if Tv=0 with v != 0 then (ST)v=S(Tv)=0 without v=0, which is a contradiction

[hard atlas]

ok the part I wanted to make clear is that you are using Tv=0 implies STv=0

but its not clear whether STv=0 implies Tv=0

(and not necessary here to think about)

[desert hornet]

It doesn't have to I think

Does this look better?

[hard atlas]

it looks overly complicated

$$Tv=0 \implies STv=0 \implies Iv=v=0$$

[desert hornet]

Oh I can just cut out the middle line

[twin meteor]

Denascite

[hard atlas]

or similar

[desert hornet]

looks good now

[hard atlas]

yes

[hard atlas]

btw it does cause STv=0 implies v=0 implies Tv=0. however, Sw=0 doesnt have to imply w=0 so there is definitely cause to be careful

[desert hornet]

Oh right

the kernel of ST is {0} but that doesn't mean the kernel of S is T(0)

[hard atlas]

so what you wrote earlier isnt wrong. I just personally think its a bit hard to read, its not clear what the implications are that you are using etc

and I hope you agree with me that the updated version is much clearer

[desert hornet]

Yeah i got confused while writing it too

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[slate birch]

does anyone have any theory on the way to solve a matrice with either Cramer, Rouche or Kronicker Kapelli?
because i completely forgot when a matrix is uniquely determined, undetermined or anything else, and how to actually solve them.
So help me please!!!!

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@slate birch Has your question been resolved?

[dense cedar]

You mean matrix equations ?

[slate birch]

yep yep

the onlly thing i know is Cramer because its the easiest

all you have to do is to find x, y and z with the formula if det A is not 0

but then what if det A is 0

[dense cedar]

I see

[slate birch]

or what if we have a parameter

[dense cedar]

You can learn the gaussian elimination

[slate birch]

the thing is, i just need to see some theory on Rouche and Kronicker kapelli, or at least an example, because i cannot find anything relevant online

[dense cedar]

Thats basically we generate an identity matrix by stepwise row operations

[dense cedar]

Wait one min

https://brilliant.org/wiki/gaussian-elimination/#:~:text=Row reduction is the process,is a much easier task.

https://www.geeksforgeeks.org/how-to-solve-a-system-of-equations-using-inverse-of-matrices/#solving-equations-with-inverse-matrices

[slate birch]

ok i see

[dense cedar]

Jump to the part, "solving eqns with inverse of a matrix"

[slate birch]

solve them with the inverse

hmm

i bet its useful

but can i give you an example of an equation, and you maybe try to solve it/

?

i get that you can probably solve it with the inverse of a matrix

seems pretty clear

but i was interested specifically in solving them with either Crammer, rouche or krocker capelli

[slate birch]

u there?

[dense cedar]

Hi

[dense cedar]

Okk I see !

[dense cedar]

krocker capelli is the same as the matrix inversion method !

[slate birch]

no no it isnt

i just checked in a book i have

[dense cedar]

What does it say ?

[slate birch]

so kronecker kapelli sais that if rang A = rang A exted then the system is compatible and determined

if rang A different than rang A exted its incompatible

[dense cedar]

Rang stands for ?

[slate birch]

idk how to explain Rouche in english but ye

[dense cedar]

Oh you mean rank !

I got it !

[slate birch]

the biggest matrix that you could make out of that matrix that has det = 0

and with rouche they all have to be 0 i think

im not sure

[dense cedar]

But that tells us if solutions exist or not

That's not a method to solve the system of equations

[slate birch]

solving it with the inverse of a matrix is the easiest type of exercises

and then theres also Gausses method/ row echelon form method. But its rarely used here

[dense cedar]

Gaussian is better for higher order matrices !

Then there's also the step escalator method!

[slate birch]

yea, to actually solve it its easy, you just apply some formula or do some easy method

the hard part is to show that its either uniquely determined, compatible and undetermined, or incompatible

thats the hard part for me at least

[dense cedar]

Cramers the best way ig !

[slate birch]

cramer only works on matrices with m columns and m rows

it also only works if det A diff 0

so its really rare to be able to use cramer

cramer is the easyest BY FAAAAAAAAR anyway

[dense cedar]

That's only when you have unique solutions.

[slate birch]

exactly

thats when you use cramer

[dense cedar]

To determine the rank you can convert the matrix to echelon form !

[slate birch]

and you just use x = det x / det A ; y= det y / det A

[slate birch]

dude you keep saying stuff that doesnt correlate with one another

like, sure you can do that, you can also find it another way

[dense cedar]

Rank is necessary for krocher caelli

[slate birch]

and with crammer you might be able to say whether its determined or undetermined

[slate birch]

oh yea kronecker capelli, my bad

ok i think i actually get it now

i dont think i have everything down yet

but ill just begin doing some exercises

cuz thats how you learn the theory the best

anyway, thanks a lot

cya

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[slate coyote]

hi, tried this question and i've no idea how to solve it on my own. What problem solving methods apply to this and what was your thinking behind it
,tex $\int\frac{1}{\cos x + \sqrt{3} \sin x} , dx

[twin meteor]

Rambo
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[hollow kite]

hello

divide by 2 in both the numerator and denominator

[slate coyote]

hmm alright

i don't see how you arrive at that answer

[wicked tundra]

experience..

[hollow kite]

in the denominator you'd get 1/2 cosx + sqrt(3)/2 sinx

[slate coyote]

could you please explain your thought process? i'd be grateful

[hollow kite]

which is

[wicked tundra]

tbh the more questions u solve the quicker u can get such ideas

[slate coyote]

ahh

[hollow kite]

sin(pi/6)cosx + cos(pi/6)sinx

= sin(pi/6 + x)

[slate coyote]

yo mean looking at the solution for the tricks?

[wicked tundra]

wdym by looking at the solution

u never learn mathematics just by looking at solution

u need to do it on ur own to get a hold on it

[slate coyote]

how did you solve it tho? what was the thought as in how did it strike you?

[wicked tundra]

remember the forms

sinacosb + cosasinb

[slate coyote]

"experience" here implies you need to be exposed to the probleme

[hollow kite]

it's something teachers love to give out to students

when doing trigonometry

[wicked tundra]

experience implies having solved a lot of questions

[hollow kite]

when i had my trig class, this form was something that you could NOT skip

this was a guaranteed on an exam

[wicked tundra]

one of the most important formulas in trigo yeah

[slate coyote]

uhm... that basically means you are exposed to methods

[hollow kite]

they make you notice each form

[wicked tundra]

sort of

[slate coyote]

never learnt it, strange

[wicked tundra]

ur learning integration but u didnt learn trigo yet?

[hollow kite]

idk you just gotta learn to notice

[wicked tundra]

strange

[hollow kite]

sometimes the "1" should even be written/noticed/said out loud while doing problems like these

[slate coyote]

so its an idea that you need to be exposed to? i mean the question seems quite aimless and you have no way of intuiting the solution/method

[hollow kite]

to get the hang of it

[slate coyote]

that's not what i said lol

i said this form wasn't explicitly taught

[hollow kite]

well, for me, our teacher loved to give such a question on an exam or school work, so the more you do it, the more you understand and the more you notice

[wicked tundra]

u might wanna remember these

[slate coyote]

I see, so its like an exposure thing

[hollow kite]

you might also want to remember these hell formulas

[slate coyote]

oh they can be derived from the already learnt formulas

[slate coyote]

yeah ik these

can the question be solved through other methods??

[hollow kite]

yes, as you can see, 1cosx + sqrt(3)sinx is basically somewhat in the form of sin(a)cos(x)+cos(a)sinx = sin(a+x)

so you gotta learn to understnad

[slate coyote]

i can for example, identify two aims in it:

1. remove root3
2. simplify the denominator

[slate coyote]

yes i got it when you asked me to multi and div by 2

[wicked tundra]

how are u supposed to rmeove root 3

[slate coyote]

it could also be cos(x-a) form

[hollow kite]

divide by 2

it'll be sqrt(3)/2

[wicked tundra]

yeah it can

[hollow kite]

you can remove that

by writing it

as cos(pi/6)

[hollow kite]

yes

[slate coyote]

so one idea that i can extrapolate from that is that always try to convert nums to values of trig functions

that way i can get a trig function to replace the mess

and then i'll be able to see the sum/diff formulas

[hollow kite]

Expressions of the form $acosx + bsinx$ can be rewritten in the form of $Rcos(x-phi)$, where $R = \sqrt{a^2 + b^2}$and $tan(phi) = b/a$

[twin meteor]

alex <3

[slate coyote]

is there a name for this method? where did u get this from?

[hollow kite]

I don't know if it's a method, it was something I was taught as well as a method.

[slate coyote]

nice, i guess i'll close the discussion here. Thanks alex!

.close

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[hollow kite]

you can also use another method, typically used in integrals, it's called the tangent half-angle substitution!

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oops

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[chilly locust]

What is the precise definition of cardinality? My searches on google doesn't give me a direct one, but just what two sets having the same cardinality means.

[gaunt sparrow]

Usually it's defined such that two sets have the same cardinality iff there is a bijection between them

[vocal sleet]

@chilly locust Has your question been resolved?

[chilly locust]

To me, it feels like saying that ordered pairs are pairs that are equal if and only if their coordinates are respectively the same (but the notion of order doesn't exist in pairs). It's like you're describing their fundamental property, but not the object itself.

[gaunt sparrow]

It's defined with the bijection because it extends to any set the notion of two sets having the same number of elements, even if they aren't finite.

Also ordered pairs do exist using set theoretical definition. (a,b) = {{a}, {a,b}}

[vast shale]

for a finite set A you could say that the cardinality of A is the unique natural number n for which there is a bijection f: n -> A

by the way, this (possible) definition assumes (possibly) a von neumann construction of numbers (for example 3 = {0,1,2}). So, if A = {a,b,c}, (with cardinality |A|=n=3, you could say there is a bijection f: {0,1,2} (which is just 3) to A

f(0)=a, f(1)=b, f(2)=c, or any other choice

there might be different definitions to capture the notion of cardinality of finite set with possibly different constructions though

[chilly locust]

That's interesting, thank you

[chilly locust]

I was expecting something that way for cardinality.

I think I'll keep looking

Thank you

[gaunt sparrow]

That's kind of the set theoretic definition of cardinality though

[cobalt crypt]

one needs to realise that many things in mathematics are defined by their properties not by explicit constructions, because mathematics cares not for how they are constructed but how you can use them

[cobalt crypt]

for instance this is one specific construction of ordered pairs in set theory, but it is certainly not the only one

and in practice, nobody ever uses that construction either

[chilly locust]

I understand, but the notion of cardinality is really quite obscure to me

[vast shale]

for a set, which is what just asked, just intuitively the number of elements of a set (finite)

[cobalt crypt]

cardinality is just an equivalence relation on all sets

[vast shale]

i would think of cardinality of what you are thinking of as number of elements in the most simple way

[cobalt crypt]

cardinality is to sets just as isomorphisms are to groups/rings/fields/modules etc

[chilly locust]

Cool!

[cobalt crypt]

cardinality for finite/countable sets has the sensible interpretation of just counting how many elements there are

cardinality for uncountable things is just hard

[chilly locust]

So, the cardinality of a set A is the equivalence class of sets that can be placed in bijective correspondence with each other, right? If so... wow... how deep

[cobalt crypt]

barring set theoretic issues, pretty much

[chilly locust]

This is really impressive

I appreciate your patience

Thank you all

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[rough pine]

how can i solve part b

by using the graph

this is a non calculator question

[crude yoke]

Just draw a line y = -1/2

[silk osprey]

i mean this is just a common value you should know

[crude yoke]

and see where it cuts the graph

[silk osprey]

from the 30-60-90 special right triangle

[rough pine]

but u cant just read it from the graph and be acurrate

[rough pine]

how do i work out the exact point

it cuts it

[crude yoke]

You don’t :)

knief is right, you should remember it

[rough pine]

i know that sin30 is 1/2

thats all i can remember tho

[silk osprey]

then you remember quadrants

[half imp]

Yes, cardinals have a specific rarely-used "definition" as one particular set: the set of ordinals with cardinality smaller than k is the cardinal k

But it's easier to think of them as equivalence classes of sets even though technically those are proper classes and not sets

Ignore me current people

What quadrants is sine negative in (ig that's apparent from the graph) and what is the reference angle of 30° in those quadrants

Or if you don't use reference angles, try to use symmetry with the unit circle, or just with the graph of sin(x) maybe

[rough pine]

bottom two quadrants

and wdym reference angle of 30 degrees in the qaudrants?

[half imp]

Or what angles have a reference angle of 30 degrees I mean

Maybe it's easiest just to use symmetry though

[rough pine]

ok so

given sin30 = 1/2

acc

is sine symmtetrical by 180 degrees?

acc no

[half imp]

sin(180+x) = -sin(x)

sin(180-x) = sin(x)

[queen prism]

Any1 free to help me in mechanics?

[gritty sage]

!occupied

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[wraith ivy]

I tried putting this into multiple calculators, didn’t work then I tried ai and they all gave different answers can someone please just help me find the answer to this equation: 30+30+30+80+80+30+80+80+120+80+30+30+30+200+120+200+120+200+200+120+200+120+200+120+200+120+200+120+30+30+30+30+30+80+80+30+30+80+80+30+30+30+30+30+30+30+30+30+30+30+30+30+30+30+30+30+30+30+30+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+30+30+30+20+80+80+200+200+120+200+200+120+80+20+30+200+30+30+120+80+80+80+30+80+80+80+80+80+200+200+120+200+200+120+30+30+200+200+120+200+120+30+200+120+200+30+30+30+120+200+120+200+120+200+200+120+80+80+30+80+200+200+120+120+200+200+200+200+80+200+80+80+80+80+30+80+30+30+30+30+30+30+30+30+30+30+80+30+80+80+30+30+30+30+30+30+30+30+30+30+30+80+80+80+80+30+30+30+30+30+30+30+80+80+30+30+30+30+30+30+80+30+30+120+80+80+80+30+30+30+30+80+80+80+80+80+80+80+30+30+200+30+30+30+30+30+30+200+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+30+30+30+30+30+80+120+80+80+80+30+200+30+30+80+30+30+80+80+30+30+30+30+30+30+30+30+30+30+30+30+30+30+30+30+80+80+30+30+30+30+30+30+80+30+30+30+30+80+20+20+20+20+20+20+20+20+20+20+19+399+100+25+10

[mild flower]

python should be able to do that

[wraith ivy]

How do I get python 🙏

[mild flower]

put print(30+30+....+10) into this https://www.programiz.com/python-programming/online-compiler/

[wraith ivy]

Okay I’ll try

[sly sierra]

google agrees with matlab, for whatever that's worth

[lost barn]

What is bro calculating

[sly sierra]

chatgpt seems unable to do it, it's just sitting there idiotically (more so than usual)

it finally gave an answer that disagrees with the others

[wraith ivy]

it says its 24733

[mild flower]

where did you get this

[sly sierra]

haha

[wraith ivy]

its from transaction logs

[cyan talon]

well gpt ain't wrong, it said xe

[wraith ivy]

i had to go through like 90 pages of stuff

[sly sierra]

i had to train it to do that

[mild flower]

if you had it as a list it might be better to just put it in google sheets or something

[silk osprey]

what happened to the XE guy

[wraith ivy]

the thing is the list is full of other stuff unrelated to those numbers

[sly sierra]

python also says 24733 btw

[cyan talon]

i don't even remember their pseudo tbh, something with josiah ? I can't seem to find them w/ discord search maybe they left

[sly sierra]

yea josiah something

ah it was maybeJosiah

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[tawny nacelle]

,, e^{tD} = \sum_{k = 0}^{\infty} \frac{t^k}{k!} D^k = \sum_{k = 0}^{\infty} \frac{t^k}{k!} \begin{pmatrix}
a^k & 0 \
0 & b^k
\end{pmatrix}

\sum_{k = 0}^{\infty}
\begin{pmatrix}
\frac{t^k}{k!} a^k & 0 \
0 & \frac{t^k}{k!} b^k
\end{pmatrix}

\begin{pmatrix}
e^a \sum_{k = 0}^{\infty} t^k & 0 \
0 & e^b \sum_{k = 0}^{\infty} t^k
\end{pmatrix}

[twin meteor]

ourfallenstars

[tawny nacelle]

is this the correct computation for $e^{tD}$ if $D = $
\begin{pmatrix}
a & 0 \
0 & b
\end{pmatrix}

[atomic jasper]

what is D?

[twin meteor]

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

[atomic jasper]

$D=\begin{pmatrix}a&0\0&b\end{pmatrix}$?

[twin meteor]

Bonk

[tawny nacelle]

yeah

I fat fingered the enter before I was able to type it out

[sly sierra]

can you explain how you went from the left to the right here?

[cyan talon]

you did something like sum(ab) = sum(a) sum(b) did you not?

[tawny nacelle]

uhh, move the sum into all components

[sly sierra]

yea but how did you pull out e^a

$\sum_{k=0}^\infty \frac{(at)^k}{k!} = ?$

[twin meteor]

Bungo

[tawny nacelle]

oh yes, yes I did

that's a problem

[tawny nacelle]

e^(at), yeah?

whoops

I knew something was wrong haha

[mild flower]

the secret code of "diagonal matrices are just funny looking numbers"

[tawny nacelle]

okay lemme do some more of these

thanks everyone

.solved

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[mental egret]

Found this yesterday

matrices are crazy ._> prolly the reason why me n them don't get along

[oak magnet]

Actually this kind of product exist

Hadamard product iirc

[heavy yoke]

the meme is that the hadamard product coincides with the ordinary matrix product for these particular matrices

[oak magnet]

Hadamard product should be the matrix default product

[heavy yoke]

not really

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[tawny nacelle]

how can I decide what these series equal?

$$\sum_{k = 0}^{\infty} \frac{t^k}{k!} \sin \left(\frac{k - 1}{2} \pi \right)$$
$$\sum_{k = 0}^{\infty} \frac{t^k}{k!} \cos \left(\frac{k + 1}{2} \pi \right)$$

[twin meteor]

ourfallenstars

[tawny nacelle]

I'm a little lost

[red bear]

you can simplify the terms

and recall the taylor series for sin and cos

[bitter pilot]

You are having multiples of pi/2

so some terms will vanish, some will change sign

[tawny nacelle]

right

[bitter pilot]

so basically in the first one you will notice -1, 0, 1, 0 as a pattern before it repeats

for sine

so the odd terms vanish

and the even terms change sign (-1)^(k+1) for example

so you could replace k = 2n where n = 0 and leave sine and add instead a factor (-1)^(n+1)

[hard atlas]

and it doesnt hurt to just write out the sigma notation sometimes

that will make it much more obvious whats going on

[tawny nacelle]

I'm having difficulties reindexing

[hard atlas]

reasons why you write out the summation

[hollow kite]

ratio test?

[heavy yoke]

that just determines convergence

[hollow kite]

oh they're asking

for the partial sum

[tawny nacelle]

,, \sum_{k = 0}^{\infty} \frac{t^k}{k!} \sin \left(\frac{k - 1}{2} \pi \right) = \sum_{n = 0}^{\infty} \frac{t^{2n}}{(2n)!} \sin \left(\frac{2n - 1}{2} \pi \right) = \sum_{n = 0}^{\infty} \frac{(-1)^{n + 1} t^{2n}}{(2n)!}

this seems to be it

ah, hold on

[twin meteor]

ourfallenstars

[tawny nacelle]

uhh, this is slightly wrong, I think

that (2n)! in the denom is problematic for n = 0

oh hold on

[bitter pilot]

0! = 1

[tawny nacelle]

..

I'm an idiot

[bitter pilot]

wolfram agrees

[bitter pilot]

i mean agrees to this

[tawny nacelle]

uhh, this is...

cosine?

I've forgotten the Taylor series

lemme check

oh it's -cosine I think

[bitter pilot]

yes

[tawny nacelle]

then the other two should be sin and -sin

let me check

[bitter pilot]

I mean both series are related in a sense that one is the derivative of the other

vaguely

[tawny nacelle]

,, e^{tJ} =
\begin{pmatrix}
\cos t & \sin t \

• \sin t & \cos t
  \end{pmatrix}

[twin meteor]

ourfallenstars

[tawny nacelle]

that's what I'm getting

[bitter pilot]

yea this is now out of my expertise

[flat whale]

How do the series relate to J

[tawny nacelle]

see hlounge

[flat whale]

Yea that closely matches what Ann said

Up to a sign. Not sure if that's her mistake or yours

[tawny nacelle]

well, that's not a large issue

I can check the sign again

thanks adonis, riemann, Bair, cloud, Ann, Denascite for the help

.solved

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[flat whale]

,w exp({{0,-t},{t,0}})

[twin meteor]

Results provided by WolframAlpha

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

Requested by rie.mann

[tawny nacelle]

flip

always the minus signs

[flat whale]

Reindexing bain of your existence

[tawny nacelle]

[flat whale]

Reindex your existence so the version of you that knows reindexing is now

[tawny nacelle]

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[lilac pebble]

for part b the answers say 6,246 but i think it's a typo

i got 6,426

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@lilac pebble Has your question been resolved?

[quiet echo]

!show

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Show your work, and if possible, explain where you are stuck.

[lilac pebble]

it uses burnside's formula

[quiet echo]

[lilac pebble]

the group is D_4

[quiet echo]

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@lilac pebble Has your question been resolved?

[sudden compass]

a shud be 720 i think? and for b...6^6 🤣

[little cedar]

Because we can use rotation I don't think it will be distinguishable

[sudden compass]

Ik my answer to b is wrong I am thinking how to get the correct answer without whatever formula that is

[little cedar]

Ok I did found that

Uhh

How bout the rotation of the 4 long side

Alr alr what is the logic behind question b

How to make it distinguishable

[lilac pebble]

the first one is 90

720 / 8

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[severe stone]

i only have fn+1 (x) = (fn(x))² i have no idea what to do

[cyan talon]

just compute a few functions

what do you think f_1, f_2, f_3, ... are?

[severe stone]

no idea

idk really what the subscript will do to the function

[ebon rapids]

can you at least try them

[cyan talon]

if you plug in n=0 here what do you get ?

[severe stone]

ooooh

f0 o f0

[ebon rapids]

which is

[severe stone]

x^4?

[cyan talon]

yeah and the left side is ?

[severe stone]

f1

so with that

[cyan talon]

yea

[severe stone]

i can calculate

f2

etc etc

[half imp]

yes do that

[cyan talon]

yea it's still a recurrence relation

[severe stone]

aaah okay

[half imp]

see if there's a pattern

[severe stone]

thank you guys

.close

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[worldly relic]

please i need help to proof the Reverse Implication

.close

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[sour valve]

Hello, I can't seem to understand the last line? I'm not sure where the 9 has come from because when I was working on the question I had 3 instead of the 9. Please can someone help me.

[versed bane]

did you check with small examples to see if your statement is actually true?

33-6 is 27

[sour valve]

ohh

[sour valve]

I'm not sure because there was a solution to the question.

[peak matrix]

the proof is correct exactly up to the point where they use 9 instead of 3. That's the mistake that makes it wrong

[sour valve]

so the solution is wrong??

[halcyon ice]

yeah

[sour valve]

ohh okay

thanks

[halcyon ice]

its supposed to be (66...66)^2 x 3

i dont know how you would go about proving it from there tbh though

[peak matrix]

this essentially disproves it

and proves that no such number can be a square

what's perhaps interesting is that if we divide it by 3, then it's gonna be a square

[halcyon ice]

wait so if m = 3n^2 then m cant be a square number?

[peak matrix]

so
11...11 - 22...22
(2n) (n)
is gonna be a square

[peak matrix]

correct

[halcyon ice]

good to know

[peak matrix]

sqrt(m) = sqrt(3) * n which cannot be a whole number

[sour valve]

sorry where did the 11...11 come form??

[peak matrix]

sorry, i made a typo. I divided it by 3 though

[sour valve]

im not sure it was just a question i got from a book

[halcyon ice]

oh right
well thats the shortest proof by contradiction ive ever seen

[sour valve]

ohhh okay thanks

[vocal sleet]

@sour valve Has your question been resolved?

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[vocal sleet]

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[mild trench]

can someone help me with 4a+b and 5a

i dont know how to draw the angle on which point of the triangle

[stone gazelle]

The angle needs to be adjacent to the positive x axis

[mild trench]

could u aid me with a diagram

[mild trench]

how do u know it goes from the axis upwards

why cant it be that?

[stone gazelle]

It can be

The important part is to make sure you take care of the sign

Counter clockwise is positive

Clockwise is negative

Say the yellow angle is +pi/4 rads

[mild trench]

i havent learnt that

[stone gazelle]

Radians?

[mild trench]

yes

[stone gazelle]

Ok, say it’s +60degrees

[mild trench]

okay

[stone gazelle]

Then the red angle would be -300degrees

[mild trench]

yes

[stone gazelle]

But it is preferred to keep degrees positive, which is why I used the yellow anglw

[mild trench]

i see

[mild trench]

could you do b as well?

[stone gazelle]

No, you do b

[mild trench]

alr

[stone gazelle]

I can check it

[mild trench]

wouldnt it be more suitable to do the green arrow?

[stone gazelle]

Yeah I would to be honest

[mild trench]

mark scheme does this

[stone gazelle]

It seems unclear what the question wants exactly, but it looks like this is just a trig exercise

So keep all angles positive

[mild trench]

its finding vector form

using trig

[mild trench]

would the clear answer be -53.1?

because the person that did it went clockwise

[stone gazelle]

Stick to positive for now

[mild trench]

got it

[stone gazelle]

When you learn radians you will probably see why negative is useful

[mild trench]

i see

[mild trench]

shall i always mention if it is below or above the axis?

[stone gazelle]

Yeah you should

[mild trench]

got it

part a for 5

should i just do it to the right?

the one in red

[stone gazelle]

Yes that will be easier

And use clockwise/counter clockwise from j

[mild trench]

got it

i asked this question yesterday so im wondering how u will do it if u dont mind

bc i always see so many methods to people doing this question:

[stone gazelle]

Well first, what would a random vector parallel to i + j look like

[mild trench]

one line will be the multiple of the other but go in the same direction?

[stone gazelle]

Yes, so how would you write that in maths

[mild trench]

k(i+j)=i+j

where k is some multiple

[stone gazelle]

Yes

[mild trench]

can i use the word scalar?

[stone gazelle]

Yes, k is a scalar

[mild trench]

alright

[stone gazelle]

Now, you are told c + ld is parallel to i + j, meaning it will look like k( i + j )

[mild trench]

yes

[stone gazelle]

So you have

c + ld = k( i + j )

[mild trench]

yes

[stone gazelle]

I would write that out in column vector form

Then just combine and compare

[mild trench]

would it just be an equation

[stone gazelle]

It’s not really an equation, more of an identity because you want the left hand side to look like the right hand side

But combine the left side into one vector

[mild trench]

done

[stone gazelle]

Now you want the top and bottom of that vector to match so that it looks like k(1,1)

So you equate them

Then solve for l

[mild trench]

i see

that makes sense

i use to do this method but i had no idea how to do this question:

beacuse k would be 0 for j

[stone gazelle]

I’ll use x for lambda because l looks odd

(2 + 3x, 5 - x) should look like (k, 0)

[mild trench]

yes

[stone gazelle]

Then I would start with the bottom, 5-x=0 so x must be 5

[mild trench]

yes

[stone gazelle]

That’s all there is to it

[mild trench]

oh yh

[stone gazelle]

Because it doesn’t matter what the top is, it will always be a multiple of 1

[mild trench]

ive seen this method as well:

[mild trench]

i see

[stone gazelle]

Just write everything out in column vector form, makes things easier

I never liked the ai + bj form

[mild trench]

ok tysm, i got 1 more question regarding what it means by "and i"

it it just asking along the x axis?

[stone gazelle]

Yes

i is 1 unit along x axis

j is 1 unit along y axis

So the angle between AB and i is the angle between AB and x axis

[mild trench]

i see ty!

[vocal sleet]

@mild trench Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[fluid flare]

,tex prove that 121 doesn't divide $n^2 + 3n + 5$

[twin meteor]

TRIAENGLE

[fluid flare]

i am unable to progress further

[quiet echo]

proof by contradiction

claim that it does divide it and provide a counterexample for n=1

(i am assuming $n\in\bN$)

[twin meteor]

;(

[fluid flare]

will proving through 11 work

[half imp]

,w does 121 divide 9?

[quiet echo]

what

[twin meteor]

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Requested by a_mog

[quiet echo]

i...don't think you need to use WA for that.

[half imp]

hmm, doesn't seem to work

[quiet echo]

?

[half imp]

Oh

For all n obviously

You can't just show it doesn't divide for one n and then be done

[quiet echo]

.

[half imp]

No

[quiet echo]

what does this mean?

[fluid flare]

i tried it by showing that 11 doesn't divide the expression

[half imp]

Oh, that would work if it was true

,w n^2+3n+5 mod 11 for n from 1 to 11

[twin meteor]

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Requested by a_mog

[quiet echo]

just do the easier and more comrephensible route

[fluid flare]

i got to n(n+3) congruent to 6 mod 11

[half imp]

n=4 though

[quiet echo]

considering that n could be dividible by 11

but not another 11

i.e., 11^2

[half imp]

My guy wtf are you on about

Everything you've said here has made zero sense

[quiet echo]

how so

[half imp]

Banana puppy ice cream

[fluid flare]

???

[half imp]

That's how so

[quiet echo]

okay...

so you wanna troll...?

[half imp]

Anyways n=4 makes it divisible by 11 sadly

[quiet echo]

i clearly stated the route here

[half imp]

so that method won't work

[fluid flare]

Oh

i will try with 121 only

[quiet echo]

.

[fluid flare]

contradictiom right ?

[quiet echo]

👍

its the most rigorous, easy route to go by

[fluid flare]

i am new to this

[quiet echo]

yeah, its fine

contradiction is a more logical approach to proving a statement "an infinite amount of ___"

[half imp]

Now you only need to test the n values that are 4 mod 11 though

At least

[half imp]

My guy you can't explain to someone why their words make no sense

That's the default state with any random amalgamation of words

[quiet echo]

??

now im more thoroughly confused

i asked for clarification on things i couldn't understand; i don't see the problem here

i gave a clear route as to how to prove it

[half imp]

You did not

[quiet echo]

[half imp]

You gave a route to prove that there is some n value which makes it not divisible

That is not the question

[quiet echo]

i am leaving because i have better things to do

[half imp]

Better things then disrupting a help channel by not understanding the question and blocking attempts at help from someone who actually knows what they're doing?

I should hope so

[vocal sleet]

@fluid flare Has your question been resolved?

[sudden compass]

hmm, my guess would be to consider $n=11k+r$ for some $0 \leq r \leq 10$

[twin meteor]

rak³en

[sudden compass]

From there's theres probably some mod shenanigans that'll work

[sudden compass]

11 does divide the expression, so this approach is a dead end

[sudden compass]

contradiction is possible too ig

[mental egret]

Use induction :| I believe it's an induction exercise. Right? m I tripping

[sudden compass]

LOL I thought that too but I doubt thats possible

I've never done a question like this using induction

[mental egret]

XD

[sudden compass]

I guess you would assume $n^2 + 3n + 5 = 121k + r$ for some $r \in \left[0, 10\right]$

[mental egret]

=> 11 | n - 4 
=> n = 11k + 4```

[twin meteor]

rak³en

[sudden compass]

where did n go

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[mental egret]

this would work

[mental egret]

@fluid flare

[sudden compass]

thats n^2 - 8n + 5, not n^2 - 3n+5

[mental egret]

yes, which is n² + 3n + 5 - 11n

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[mild trench]

if i integrate pq

will it be px qx?

[wraith mist]

pq is just some constant A. What do you get after integrating A?

[mild trench]

Ax

[wraith mist]

Yes

[mild trench]

how do i know that

pq are constants

what if p and q are y and x

idk if im making sense

[mild flower]

because it asks for the value of p and q

[mild trench]

oh ok

[wraith mist]

Yeah p and q are just unknown numbers

The goal is to find them

[mild trench]

i need help with the wording of this btw:

second line

why do i need to know that x is small

[mild flower]

so that you can ignore terms that involve x^7

[mild trench]

can u use an example of what u mean

[mild flower]

so if you have x = 0.1 for instance

[mild trench]

yes

[mild flower]

then x^3 = 0.001 which is too small to think about

[mild trench]

ohhhhhhhhhv

ok

ty

one last question

[mild flower]

and ofc x^4 is even smaller than that

[mild trench]

it asks for the exat vaule of k

however

in the mark scheme it says

+/- 6

but it says value, not vaules

so like

doesnt really make sense coz shouldnt k have only one value, not 2?

[quiet echo]

well, then thats an error on their part, i suppose

because both of those values are valid

[mild trench]

ohhhhhh ok ty

[quiet echo]

let me just double check

[mild trench]

okay

[quiet echo]

$\sqrt{4+k^2}=\sqrt{40}$

aa

[twin meteor]

;(

[quiet echo]

i jumped ahead and behind myself sry

[mild trench]

its ok

[quiet echo]

$k^2=36\implies k=\pm 6$

[twin meteor]

;(

[quiet echo]

yeap

[mild trench]

right

[quiet echo]

they're both valid

[mild trench]

i see

ty

[vocal sleet]

@mild trench Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[vocal sleet]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[misty goblet]

hi, how can I find the limit of $(\sqrt[n]{n} - 1)^n$ using the squeeze theorem? I thought about using $\frac{1}{n}$ and $-\frac{1}{n}$ but not sure how to show they are the "bounds"

[twin meteor]

p1za

[misty goblet]

e.g. proving that $(\sqrt[n]{n} - 1)^n \leq \frac{1}{n}$ or should I approach this differently

[twin meteor]

p1za

[flat whale]

You can factor out $n^{1/n}$ from the parentheses and use binomial theorem on the remaining $(1-x)^n$

[twin meteor]

riemann