help-27

[graceful stone]

It's just parentheses

[snow raptor]

floor function

ceiling function

[fading sorrel]

no its just normal

[graceful stone]

Round function no?

[vivid estuary]

hmm couldnt it be true

[snow raptor]

and others, but they state [.] - floor function

[fading sorrel]

how come?

[graceful stone]

Just not to confuse @fading sorrel , it's probably not the floor, ceil or round function

[vivid estuary]

Im not sure if this is right, but if g(x) is some pieceswise function not defined at some point x and f(x) is the function that fills that hole then wouldnt limit f(x) + g(x) exist but not g(x) at x?

[fading sorrel]

ye

[fading sorrel]

ohhh hmmmm

i never considered piecewise

[alpine python]

i feel like f would need to be discontinuous to fill in the hole

but idk

[fading sorrel]

but then wouldnt gx be composed of h(x) and k(x) lets say for some values of x

[snow raptor]

yea, they state it before any problem

[vivid estuary]

ahh wait

im thinking of continuity

[vivid estuary]

this is wrong

[alpine python]

wait can't you subtract

lim(f+g) - lim(f)
since they both exist you're allowed to do:
= lim(f+g-f)

i think

and that would imply lim(g) exists

[vivid estuary]

that seems reasonable to me yeah

[alpine python]

or...

that rule usually goes in the other direction

separating into 2 limits rather than combining into 1

[vivid estuary]

limg(f+g) - limg(f) = lim(f) + lim(g) - lim(f) = lim(g) is the right way to go aboue it I think

where subtraction. is closed under the real numbers which would maen lim(g) has to be real?

[alpine python]

[acoustic leaf]

lim(f+g) = lim(f) + lim(g) only applies if both limits exist individually

[alpine python]

yeah

[vivid estuary]

ohh right

[alpine python]

that's why i used f+g and f

still, maybe it doesn't work

[fading sorrel]

wait lets say gx does not have a limit at x = c
then this means that it is not continuous at x = c
and fx is a normal continous function that does have a limit at x = c

when we add fx + gx, using continuity laws we just get another discontinous function at c. what an i do from here

[analog flume]

it's incorrect to say that if it does not have a limit then it is not continuous

[fading sorrel]

isnt it true tho? for piecewise and lets say 1/x
at x = 0, its discontinuous so no limit

[vivid estuary]

if it doesnt have a limit its not continouts at x but I think he means its incorrecct to say if its not continues then it doesnt have a limit

because there are 3 rules to continuity

and one of them being broke would make it not continous but the lmit could still exist

[fading sorrel]

no limit then no continuity
however the converse is not true

[analog flume]

omg i was thinking derivative not limit ignore me

[fading sorrel]

lol all goods

honestly i need to get better at this

i just finished yr 12

in aus

[alpine python]

we don't have f continuous necessarily

[fading sorrel]

ye we cant use this if ones to infinite

so the only way to do this is to manipulate the inside of the brackets

[vivid estuary]

if the lim f(x) at x = c but the lim of g(x) doesnt exist at x = c, is it safe to assume that the lim f(x) + g(x) at c will still always exist?

[fading sorrel]

ye not necessarily

[vivid estuary]

hmmm

[fading sorrel]

i cant even use limit laws for this

they dont even work for non existant cases

[alpine python]

all right i found a theorem in spivak

[vivid estuary]

currently reading spivak lol

on chapter 2

😭

[fading sorrel]

whats that

[alpine python]

page 102 theorem 2

[fading sorrel]

ok lemme draw something

[alpine python]

If $\lim_{x\to a} f(x)=l$ and $\lim_{x\to a} g(x)=m$ then $\lim_{x\to a} (f+g)(x)=l+m$

[woven radish]

Axe

[fading sorrel]

ye thats for cases that exist

[alpine python]

they do exist if you apply it right

one function is f+g and the other is -f

[vivid estuary]

hm wait

[fading sorrel]

yooooo wait
thats smart
so lim( fx + gx -fx) = lim(fx + gx) - lim(fx)

lim (fx + gx) exists and lim fx exists hence lim(fx + gx - fx) as well
therefore lim(gx) exists

[vivid estuary]

ooo

[alpine python]

yeah basically

[vivid estuary]

smart smart

[fading sorrel]

thats really clever

[alpine python]

🧠

[vivid estuary]

its a calculus book by Micheal Spivak known for being the best if not best calculus book but also pretty rigorous

[fading sorrel]

aight ill try to get a pdf

i think we can also try to prove it using proof by contrapositive

if lim gx does exist then lim fx + gx or lim fx does not exist

but i think this will make it harder to consider cases etc

[misty crest]

why lim g does exist?

original statement was lim (f+g) and lim f exists -> lim g exists and you either show it’s true or that it’s false meaning it’s possible for lim(f+g) and lim f to exist while lim g does not exist

T -> F

so the contrapositive is lim g DNE -> lim (f+g) DNE or lim f DNE

[graceful stone]

Is there any reason we want a contrapositive proof?

[devout snow]

@fading sorrel Has your question been resolved?

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[fading sorrel]

ye it dont work

[devout snow]

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[devout snow]

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[rustic jetty]

is it true that $5^{\nu_5(n)} = n$

[woven radish]

Dork9399

[rustic jetty]

its only true iff n is a perfect power ie n = 5^a right

$\nu_5(n)$ is the 5-adic valuation of n

[woven radish]

Dork9399

[sage burrow]

what should that be?

[rustic jetty]

?

.close

[devout snow]

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[sage burrow]

nice strategy. closing instead of answering.

[rustic jetty]

i figured it out calm down

[devout snow]

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[arctic pendant]

help

[arctic pendant]

This is an attempt to prove the integrability of the function defined

Here’s my my attempt

If there’s a formatting issues please point out so I can improve

[devout snow]

@arctic pendant Has your question been resolved?

[arctic pendant]

<@&286206848099549185>

[jade oak]

Why are you defining $M_k$ as the sup over the interval $[0,1]$ and not $[x_{k-1},x_k]$?

[woven radish]

謎の男

[arctic pendant]

because in some interval it can be 0

[pine bone]

ema

[arctic pendant]

Density property of Q in R

But if I let it be on [0,1] then I never have to discuss that

The chance of x_k is rational is super small though

[jade oak]

The issue is your also using the $M_k$ in the sum for $U(f,P)$, but the sum in how $U(f,P)$ is defined relies on $M_k$ being defined over the interval $[x_{k-1},x_k]$, so when you say $U(f,P) = \sum \dots$ it is not true

[woven radish]

謎の男

[jade oak]

Also the sum would still just be 1

[arctic pendant]

Can I replace it with $\sum\limits_{a_n \in [0,1]}$ for the upper sum

[woven radish]

Emmaaaaa

[arctic pendant]

I don’t think so now

[jade oak]

$U(f,P)$ is defined to be $\sum_{i=1}^n M_k(x_k-x_{k-1})$ where $M_k := \inf{f(x) \mid x \in [x_{k-1},x_k]}$ so you cannot change it at all. You need to use this definition

[woven radish]

謎の男

[arctic pendant]

Yes if I do it with other it’ll be the integral instead

[arctic pendant]

Then may I ask how I can properly discuss that there exist supremum of U(f,P) that is 1 over the sub interval

I was confused about it so I came up with that notation

[jade oak]

I don't think you need to do this at all for the proof, but if you wanted to, we know $1 \in {a_n}$ so choose $n$ such that $a_n = 1$ then argue $a_n$ is in some sub interval $[x_{k-1},x_k]$ and the supremum for that interval only will be 1

[woven radish]

謎の男

[arctic pendant]

Can I simply write M_k = sup x in [x_k-1,x_k] =1?

[jade oak]

It would only be true for 1 choice of k

but if you can figure out that k then yes

[arctic pendant]

Yes so best way is to argue there must be an N such that the supremum in that sub interval is 1

Am I understanding it right

[jade oak]

I don't think so (I personally don't see this thought process going anywhere).

Instead I would argue that M_k is always less than or equal to 1 (not just equal to) and somehow use that, but this is not the "hard" part of this problem

[arctic pendant]

Okay I got it then I will try to firstly refine it a bit… this part has been causing me fairly a lot of confusion

[jade oak]

Have you heard of dirichlet's function?

[arctic pendant]

Yes

[jade oak]

Do you know why it is special with regard to integration?

[arctic pendant]

Each sub interval contain rationals

And I also know a Dirichlet test but don’t know if it’s relevant

[jade oak]

Nah I was gonna say (if im understaning the problem right) this seems to just be dirichlets function on [0,1] so if you know why that is/isnt integrable on a given interval then you can use the same logic here

I assume {a_n} being an enumeration on [0,1] with such a_n in Q means its an enumeration of Q intersection [0,1] right?

[arctic pendant]

Yes

That’s true

That should be the idea

Omg you’re a genius that’s exactly the point

[jade oak]

Yeah I didn't realize at first and was confused trying to figure out how you would do this

[arctic pendant]

That’s super clever 🥰🥰

Now this should be proven

[arctic pendant]

With that part fixed it should be considered rigorous right?

[jade oak]

Confused what you mean. What should be considered rigorous?

[arctic pendant]

If I fix that part, the rest of the proof is fine right? Is there more I need to fix?

[jade oak]

Nah the special thing about dirchilets function is || its not riemann integrable so you need to prove why this is the case||

[arctic pendant]

Because the discontinuity in this case of finite?

I didn’t think so deep behind I was so focusing on inequality

[jade oak]

There are uncountably many discontinuities which suggests it may not be integrable but is not actually a sufficient reason

Without giving a proof its not really obvious why its not integrable

like intuitively you would think its 0 (which it is for some other types of integrals)

[arctic pendant]

I should fix the sum them

So there is 3 terms of the enumeration

Maybe it’s not integrable

[jade oak]

Ok I am realizing I made some small errors that might be confusing you a bit

[jade oak]

First here I meant to write supremum not infimum

[jade oak]

And this is actually true for all k

So M_k is always 1

BUT you still cant define it the way you did in the proof, you need to argue that the supremum is 1 over every subinterval for any partition

Once you do this you will see U(f,P) is always 1 and L(f,P) is always 0

[arctic pendant]

Yes I know get it

Than it’s divergent sum

[jade oak]

Not divergent because you are still multiplying by the length of the sub intervals which are small

In fact, knowing that M_k = 1, the upper sum is just summing all the lengths of the intervals

which is why we know the upper sum is 1 (because the partition is taken over an interval of length 1)

[arctic pendant]

But there should be infinite many rational over [0,1] right?

so summed together it’s still huge

I am super confused

Now

[jade oak]

yeah but we are basically only picking one rational for each subinterval

[arctic pendant]

Enumeration must be finite?

[jade oak]

Nah, but partitions are finite

and the upper sum is finite because of this

(another reason it can't diverge, its not a series)

[arctic pendant]

Yes I think I must watch a bit more videos, basically this part is abstract too abstract

And I feel it so hard to learn it without taking lectures

[jade oak]

Yeah

[arctic pendant]

🥰🥰🥰thanks so much anyway you helped my life

[jade oak]

np

[arctic pendant]

I fixed some parts now it should work right?

I finally got what you mean by the partition is finite

And now the logical counter example is corrected!

[jade oak]

You should get that the function is not integrable

The first half of the proof is good

You need to also explain why $M_k$ = 1. You can do this with basically the same exact argument you did to show $m_k = 0$.

How are you getting $\sum_{k=1}^n M_k(x_k-x_{k-1}) = \frac{\epsilon}{\text{tree}(3)}$ though. Remember $M_k = 1$ thus $\sum_{k=1}^n M_k(x_k-x_{k-1}) = \sum_{k=1}^n (x_k-x_{k-1})$ for any partition. This is still computable though.

Consider a small example. What happens if you have the partition ${0,0.25,0.5,0.75,1}$ (here n = 5)

Then $\sum_{k=1}^4 (x_k-x_{k-1})$ = \sum_{k=1}^4 \frac{1}{4} = 1$. The same logic will apply to any partition: the sum of all partitions over an interval is the length of the interval.

[woven radish]

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

[arctic pendant]

But there is a distinction between this and a dirichlet function, that one has infinite many discontinuity

And since Q is dense in R so every sequence of Q contains elements in real like right?

since its one enumeration of points and the interval is so small

We are basically integrating 0

[arctic pendant]

If it’s not integrable then I am even more confused because dirchlet is not integrable bcs infinitely many discontinuity

[jade oak]

I think I might be confused on the function. Is it not true that for any rational number a_n in [0,1] we have $f_j(a_n) = 1$

[arctic pendant]

In my case it’s finite

[woven radish]

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

[arctic pendant]

Yes

But enumeration is finite right

It has n terms

[jade oak]

Not necesarrily

[arctic pendant]

If it’s finite then it’s integrable right?

But this one had n term right?

[jade oak]

Ok there is maybe a problem of notation

You are using $a_n$ to describe elements in the enumeration

[woven radish]

謎の男

[jade oak]

but you also are using $n$ to be the number of points in your partition

[woven radish]

謎の男

[jade oak]

is this correct?

[arctic pendant]

Yes

True I made some confusion with that

[jade oak]

Yeah they are separate things

I assume your enumeration is over the rational numbers in Q. Even if it just means some subset of Q, in the worst case it could still be all of Q

thus your enumeration is infinite

[arctic pendant]

I will refine that part too! If a_n is finite then its integrable right

[jade oak]

Yeah

[arctic pendant]

I am actually not studying mathematics so my notation is super messy😭😭

I will be fixing it right now

[jade oak]

Tbh I think im very confused on the set up of the problem now.

I was originally thinking that ${a_n}$ was an enumeration of ALL rational numbers in $[0,1]$, but you are saying that ${a_n}$ is finite. Is it actually a finite enumeration of SOME rational numbers in $[0,1]$

[woven radish]

謎の男

[arctic pendant]

Usually is infinite enumeration denoted like this ${a_n}^\infty_{n=1}$

[woven radish]

Emmaaaaa

[jade oak]

I think it is often just written as ${a_n}$

[woven radish]

謎の男

[jade oak]

but I may be wrong tbh

Let me relook at your proof assuming it is finite now

[arctic pendant]

I am super bad with these details too because I am not professionally trained 😭😭😭

[jade oak]

Yeah I self studied this so I may not be the best with notation either

[arctic pendant]

x_n should be a_n right?

[jade oak]

Nah the x_n would be a point in the partition, while a_n would be a number in the enumeration

[arctic pendant]

Then it should be f_n(x) instead of f_j(x)

I got this problem from internet 😭

[jade oak]

Assuming ${a_n}$ is finite, then it is not true that $M_k = 1$ for all $k$. It may also be the case that $M_k = 0$ (I can provided an example if needed). This does not impact the proof too much though.

The main issue is I still don't get how you have $\sum_{k=1}^n M_k(x_k-x_{k-1}) = \frac{\epsilon}{\text{tree}(3)}$

[woven radish]

謎の男

[arctic pendant]

Because there are n terms ?

X_n is defined as x_{n-1} + epsilon/ntree(3)

[jade oak]

Each $x_k - x_{k-1} = \frac{\epsilon}{\text{tree}(3)}$ (where $M_k = 1$) though right?

[woven radish]

謎の男

[arctic pendant]

Yes

No

It’s epsilon/ntree(3)

The difference

[jade oak]

oops I did not see that

Yeah ok now your reasoning makes a lot more sense to me, but it is still due to the issue of using $n$ for two different things

[woven radish]

謎の男

[arctic pendant]

For that I will fix it 🥰🥰

[jade oak]

Lets instead denote the enumeration as ${a_m}$ (using $m$ instead of $n$)

[woven radish]

謎の男

[arctic pendant]

Yes i totally agree i was so confused by this notation too

[jade oak]

Then instead you will be choosing it so $x_k - x_{k-1} = \frac{\epsilon}{m\text{tree}(3)}$ (because $n$ doesnt exist yet)

[woven radish]

謎の男

[arctic pendant]

Yes that’s then correct I think?

[jade oak]

This means now $U(f,P) = \frac{n\epsilon}{m\text{tree}(3)}$

[woven radish]

謎の男

[jade oak]

If we choose a very large partition this will become arbitrarily large though

[arctic pendant]

Should I say m > n

Then it will be weird though not every interval might have rational

[jade oak]

You can't say that because we are trying to compute for a general partition, where $n$ can be arbitrarily large

[woven radish]

謎の男

[arctic pendant]

Yes I got it, epsilon is arbitrary it’s fine

I have a tendency to make it smaller even if it’s arbitrarily small already bcs I am not good with it so 😭😭

So I will fix the index now 🥰

[jade oak]

replaced with vvvv

This reason is actually bad. The actual reason is if we partition the interval $[0,1]$ such that the sub intervals have length $\frac{\epsilon}{m\text{tree}(3)}$, then we must have that $n > m\text{tree}(3)$.

[woven radish]

謎の男

[jade oak]

assuming epsilon < 1

[devout snow]

@arctic pendant Has your question been resolved?

[devout snow]

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[lusty wing]

how to easily divide a number by a decimal numberÞ

[lusty wing]

like

52*5,91

[orchid sierra]

Divide or multiply

[orchid sierra]

That's multiplication symbol

*

Bit when you do division woth decimals, you can adjust the problem by multiplying the divisor and the dividend by a power of 10 such that all decimals are eliminated

For example, suppose you are doing

70.1 ÷ 17.525

       ________
17.525 ) 70.1

Multiply both numbers by 1000

       ________
17,525 ) 70,100

Now there are no decimals.

Maybe this wasn't the best choice because that goes in exactly 4 times, but you get thr idea.

[lusty wing]

okey

but multiply?

<@&286206848099549185>

[devout snow]

@lusty wing Has your question been resolved?

[devout snow]

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[lusty wing]

.

[granite anvil]

Can someone explain what is happening in the next step:

like what is happening here too

This is the result

and then they do this

[neat halo]

what grade maths is this

[granite anvil]

prob first year uni

[neat halo]

seems easy

[granite anvil]

i am grade 12 but this is part of my math ia

yeah could you explain it

i haven't covered matrices

[neat halo]

why

[granite anvil]

like the whole point is to find a single function that will pass throuhg all the points

[granite anvil]

wdym

[neat halo]

do u find this hard

[granite anvil]

yeah i dont know how to do it?

[neat halo]

i dont know how to do it i'm too young

[elfin shadow]

Wait is it asking to reduce those matrices to echelon form

[granite anvil]

yeah

[elfin shadow]

Oh

[granite anvil]

but im not sure what is happening in the step before either

[elfin shadow]

I mean it’s basically just doing arithmetic

Kinda boring

And tedious

[granite anvil]

like how is she making this matrix

[acoustic leaf]

the matrix consists of the coefficients of each variable

[granite anvil]

yeah

[elfin shadow]

I mean yeah she’s just constructing a coefficient matrix

But like

[granite anvil]

no i mean i understand the relationship beteween each row, its just teh coefficients of abcd

but when shes multiplying the matrix with abcd is she multiplying every value with a in that row?

im not sure how the multiplication works if its not with just a scalar value

[acoustic leaf]

we define matrix multiplication in such a way that multiplying the matrix of coefficients with the vector (a,b,c,d) as shown results in a vector which is the left side of each equation

[elfin shadow]

Well she multiplies matrices

[granite anvil]

okay so what are they multiplying in that step

[acoustic leaf]

in each row, you multiply the first element in the row by the first vector element (a), then the second element in the row by the second vector element (b), and so on

then add them up

[elfin shadow]

Yeah

[granite anvil]

oh ok

yeah so its the same as the polynomial equation

that makes sense

so wahts the point of the reduced row echelon format

[acoustic leaf]

the reduced row echelon form will give a matrix, which when translated back into a system of equations, will have the same solutions but much easier to see

[granite anvil]

so basically it gives you the function that satisfies all 4 equations?

[acoustic leaf]

so if you do the RREF you will get the matrix: [ \begin{bmatrix} 1 & 0 & 0 & 0 & -0.113 \ 0 & 1 & 0 & 0 & 0.690 \ 0 & 0 & 1 & 0 & -0.882 \ 0 & 0 & 0 & 1 & 1.25 \end{bmatrix} ] which if we translate back into the equation is [ \begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0\ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a \ b \ c \ d \end{bmatrix} = \begin{bmatrix} -0.113 \ 0.690 \ -0.882 \ 1.25 \end{bmatrix} ]

[woven radish]

cloud

[granite anvil]

how does the back translatoin work?

[acoustic leaf]

to do the rref, we combine the coefficient matrix with the vector representing the right side of each equation. that's how we go from the equation shown here to the matrix actually entered in the calculator

in the calculator you then use a function which generates the RREF of that matrix. but in the RREF the last column still represents the right side like before

[granite anvil]

so (0.474)^3 x 0.973 = 0.106496?

thats not what im getting though

wait no im confused now

[acoustic leaf]

where is this calculation coming from?

[granite anvil]

correct me if im wrong

[granite anvil]

but aren't you multiplying the matrix on the far left here with the one on the RHS ?

[acoustic leaf]

no, you aren't

[acoustic leaf]

you are putting the matrix on the left and the vector on the right into one big matrix (what we call an augmented matrix) where the first 4 columns are from the left and the last column is from the right (shown in the calculator here)

you are then finding the RREF of this augmented matrix (this can be done by hand but it is easiest to make the calculator do this)

[granite anvil]

oh i see

ok ok that does make sense

so how are they finding the RREF

like if you could explain just one step of it

[granite anvil]

cuz it looks like they are just evaluating the matrix

here

idk if thats the right terminology

but they are expanding the brackets

[smoky nimbus]

Do you know how to do elimination with equations? That's pretty much what is done

[acoustic leaf]

once you have the matrix evaluated you would then have the calculator perform an inbuilt 'rref' function

[granite anvil]

if i was writing a math lab on something involving this do you think I'd have to explain how the RREF works?

at the grade 12 level

[granite anvil]

like with linear systems?

[acoustic leaf]

internally the calculator would be performing row operations (adding rows to other rows, multiplying rows by scalars, and swapping rows), but all we really need is the end result (which is the top matrix here)

[smoky nimbus]

Yes, solving a system of equations using elimination

[granite anvil]

oh yeah

[acoustic leaf]

the elimination method for linear systems is exactly how the calculator performs RREF (it just uses matrices to express the system)

[granite anvil]

so you guys think its okay to just say find RREF using graphical display calculator on a math lab?

so i dont actually have to explain how it works or nah

[acoustic leaf]

if solving a linear system of equations is a step you need to do and you are allowed a calculator that does it then you can probably use it

the exact guidelines are up to your teacher, though

[granite anvil]

oh okay ill ask

thank you so much

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[still heron]

someone pleasee pleasee help me w this

[still heron]

pweaseee

does any1 know?

<@&286206848099549185>

if any1 understand this plz lmk

❤️

[neat halo]

this cant be real

wtf is this

no one in the world does this math

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[last hornet]

Hello

[solid osprey]

hi do you have a question?

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[lost laurel]

I'm trying to show that if $(b_n) \to b$ then $(\abs{b_n}) \to \abs{b}$
\
\
Proof:-As $(b_n)$ is convergent, it follows that $\forall \varepsilon >0, \exists N>0$, st, if $n > N$, then $\abs {\abs{b_n} - \abs{b}}\leq \abs{b_n - b} < \varepsilon$
From this we can conclude that $(\abs{b_n}) \to \abs{b}$

[pastel pasture]

What's the third line for?

[lost laurel]

I need $\abs{\abs{b_n} -\abs{b}}< \varepsilon$

[woven radish]

ƒ( wai ina teacup)= I don't know

[pastel pasture]

you already have that in the second line?

[lost laurel]

oh right

well, $\leq$

[woven radish]

ƒ( wai ina teacup)= I don't know

[lost laurel]

but yeah, minor typo

[pastel pasture]

also st and if and shouldn't be in math mode

if you want to emphasise you should use italics or slant

[woven radish]

ƒ( wai ina teacup)= I don't know

[pastel pasture]

sure

[lost laurel]

Thanks!

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[woven radish]

𝔽𝟜𝕟𝕥𝟘𝕞 | Retired
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✅

[brittle burrow]

• you're told they're all positive numbers, so what the solution is saying is the smallest number possible for x1 to x10 is the smallest positive number (1)
• x11 is the median, so all the numbers after that have to be bigger than or equal to the median (10)
  so the smallest x11 to x21 can be is 10
• but then it says the biggest number was 16, so we can swap x21 to a 16 to fix that

that gives us 1 10 times from x1 to x10, 10 for the median x11, and 10 10-1=9 times from x12 to x20, and 16 for x21

yes

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[fervent ore]

How does this equal 0.94, ill post my steps below where i get the wrong answer

[fervent ore]

i get like 5.37

which is way wrong

[granite arch]

u forgot to integrate this?

[fervent ore]

Oh yeah i missed that

Lemme try again haha

It worked now thankfully!

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[regal berry]

(2x+5)[3(x-2) -2(x-4)] +2(2x+5)

I can see that at -2(x-4) 4 = -2 *-2 and to take out the common factor since we already have a number do we multiply 3 with -2 i am a little confused on the steps here

[ancient sluice]

First simplify

That will be easier to understand

3(x-2)-2(x-4) = x + 2

[ancient sluice]

Is this understandable?

[regal berry]

3x - 6 -2x + 8 = x +2 yeah

[ancient sluice]

So will have (2x+5)(x+2) + 2(2x+5)

We have the common factor 2x+5

So if we factor out, (2x+5)[(x+2)+2] is the result

[regal berry]

oh i see it now. My mistake in the first place is that i simplfied 2(2x+5) so i had 4x + 10

[ancient sluice]

Good ! Now yk what to do !

[regal berry]

thanks

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[floral barn]

help

[ancient sluice]

Hii

[floral barn]

omg dude

[ancient sluice]

Multiply numerator and deno with

(1-cosx)

[floral barn]

thanks for that exercise yesterday, it was so cool

[ancient sluice]

And write cos2x = cos²x - sin²x

[ancient sluice]

Glad to help !!!

[floral barn]

ight one sec

[ancient sluice]

(a+b)(a-b) = a²-b²
So deno becomes (1-cos²x) = sin²x

[floral barn]

yup

[ancient sluice]

So can ya solve it now ?

[floral barn]

i should write cos 2x

as 1-2sin^2x right?

and that should solve it maybe

[ancient sluice]

[(cos²x-sin²x)(1-cosx)]/sin²x

[ancient sluice]

Yeah write that

[floral barn]

oof

[ancient sluice]

Solved !?

[floral barn]

so i got cos2x - cos2x cosx

[floral barn]

almost

i got 2- pi/2

but i think it should be pi/2 - 2

[ancient sluice]

Lemme check

[floral barn]

ight

do it with cos2x= 1-2sin^2x btw

[ancient sluice]

,w integrate from 0 to pi/2 cos2x/(1+cosx)

[woven radish]

Results provided by WolframAlpha

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

Requested by xor_11

[floral barn]

dayum

[ancient sluice]

They say it's 3- pi

[floral barn]

then that the answer prolly

[ancient sluice]

Wolfram isn't wrong

🥲

[floral barn]

ik

[ancient sluice]

so would you send me a pic of your calculations?

[floral barn]

sure

i got 2-pi

idk how they got 3-pi

let me send you

[ancient sluice]

Yeah sure

[ripe hollow]

I also got the same

[floral barn]

we are probably just stupid

[ripe hollow]

lol

[ancient sluice]

cos2x(1-cosx) = (1-2sin²x)(1-cosx) = 1 - cosx - 2sin²x + 2sin²xcosx

Dividing by sin²x we get

(cosec²x -cosecx*cotx - 2 + 2cosx)

[sand dove]

forgot to expand properly

[floral barn]

ohhhhhhh

i did an oopsie

[sand dove]

the 1 in (1-2sin^2(x))cos(x) didn't get multiplied

anyways, when in doubt, tan(x/2) can do the job

[floral barn]

this should be right this time

almost right, there shouldve been a sin^2 x

[ancient sluice]

I shall say no

[floral barn]

the third one shouldve had sin^2 x

thats prolly the only thing

[ancient sluice]

Yeah and it would be (-ve) before 2

[floral barn]

oh

i see

so how do i solve it

in the first and third integral do i just subsitute with tgx/2?

[ancient sluice]

Nah they are direct integrals

[floral barn]

oh, the first one is -ctg x maybe

yes im right

and the third?

[ancient sluice]

Yeahh

[ancient sluice]

Third one is cosec x cot x

Integration of this gives

-cosecx right .

?

[floral barn]

i told you yesterday that nobody taught us cosec and sec stuff here so

i got no clue

[ancient sluice]

😭

[ancient sluice]

Teachers are like this :'')

[ancient sluice]

Differentiation of cosecx is -cosecx × cotx , so integration of the latter returns -cosecx

[floral barn]

oki so now the problem is just cosx/sin^2x

[floral barn]

do you have any other method of solving this?

[ancient sluice]

Wait let me see

This is the easiest one though, let me see for other ways

[floral barn]

sure

[ancient sluice]

We can do by writing cosx + 1 = 2cos²(x/2) but that makes the process very complicated

But let's see

[floral barn]

how about the tgx/2 substitution

[ancient sluice]

In numerator we have cos2x = 1-2sin²x

[ancient sluice]

Nah that won't work here we need to see the degree

[ancient sluice]

sin²x = 4sin²x/2cos²x/2

So we have (1-8sin²x/2cos²x/2)/2cos²x/2

We get 1/2 sec²x/2 - 8sin²x/2

Now recall

2sin²x/2 = 1 - cosx

[floral barn]

mhm

[ancient sluice]

So we have 1/2 sec²x/2 + 4(1-cosx)

You can integrate this now

[floral barn]

i didnt understand what you did

oh wait

dude

cosx = (sinx)'

so now we just got u/u^2

but wait cuz that wouldnt work either

because we would get 1/sinx

from 0 to pi/2

so we would have 1/0

[ancient sluice]

Ah what ?

I don't understand what you say !

[floral barn]

we are trying to solve integral cosx/sin^2x dx right?

[ancient sluice]

No

[floral barn]

thats what i was trying to solve

[ancient sluice]

I solved the whole one 😭😭

[floral barn]

all of those are fine

except the third one

[ancient sluice]

Nah if we proceed by this method there's no other direct way

Can you memorize it ?

[floral barn]

memorize what?

[ancient sluice]

Integration of csc x cot x = - csc x

[floral barn]

it just doesnt help me anyhow

[ancient sluice]

It does

[floral barn]

and these exercises dont require them necessarily

those formulas i mean

[ancient sluice]

cosx/sin²x = cos x/ sin x * 1/sinx

= cot x * csc x

You can try solving the integral the other way I mentioned

[floral barn]

i apreciate trying to help me

[ancient sluice]

From here

[floral barn]

is this integral cosx/sin^2x dx so hard to solve otherwise?

without the formula?

[ancient sluice]

That's the basic formulae

[ancient sluice]

Let me see this particularly

cos x = cos² x/2 - sin ² x/2
sin²x = 4sin²x/2 cos²x/2

So we get

cosx /sin²x = 1/4(cosec²x/2 -sec²x/2)

You can integrate this ?

[floral barn]

sure

wait

[ancient sluice]

Yes

[floral barn]

csc 0 is undefined

[floral barn]

so this

this formula of yours

doesnt help us at all

i tried it with t substitution and got something over 0 too

so it didnt matter after all

[ancient sluice]

Yes it helps 😭😭😭 you

[floral barn]

how?

[ancient sluice]

That's the basic formulae we're taught

[floral barn]

ok dude

i get it

it a formula

[ancient sluice]

Like int (cosx) = sin x

[floral barn]

but the integral is from 0 to pi/2

[ancient sluice]

A formula like that !

[floral barn]

my dude

if we applied your formula

and we got - csc from 0 to pi/2

how did that help you

cuz now you dont know what csc 0 is

do you get it now?

[ancient sluice]

Ok

Csc x is undefined at x = 0

[floral barn]

yep

[ancient sluice]

That's causing the problem I see, the best way this would be to use multiple angle formula

[floral barn]

you flexed on me with your useless formula for 30 mins just to abandon it

sorry if im being mean

[ancient sluice]

🙃

[floral barn]

now what?

[floral barn]

did i upset you?

[simple seal]

what is the original question

[floral barn]

this is the original question

[simple seal]

what have you tried

[floral barn]

i tried to aplify the fraction by (1-cosx)

which got me nowhere

so that is for sure not the solution

[simple seal]

have you tried kings rule

[floral barn]

can you let me know what the kings rule is?

i may know it, just not in english

[simple seal]

uh

basically the substitution u = pi/2 - x

[floral barn]

wait a sec

i think i kinda figured it out

if you wanna help me i could tell you my idea

[simple seal]

sure

[floral barn]

nvm i got it

i solved it

[simple seal]

nice

[floral barn]

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[signal canopy]

hey so i don't much about calculus but i have been told that in calculus dx is defined as increment in x . So my question is , does -dx denote decrement if yes then if x1 is lower limit and x2 is upper limit can i integrate f(x) from x2 to x1 by with respect to (-dx) instead of +dx is it correct?

[shut pagoda]

yes you could do that

[signal canopy]

so can i say that is the reason why we put a negative sign when we reverse the limit of integration?

[shut pagoda]

yup

[signal canopy]

so my physics teacher just integrated this from infinity to r without putting -dx and i was wondering how he got the correct answer . By the way that negative sign before the integral is due to vectors having opposite direction

[shut pagoda]

,rotate

[signal canopy]

i just wanted to verify if i am missing some concept from calculus

[shut pagoda]

do you know FTC part 2

this holds regardless of weather a<b

also imagine reversing a and b

then the left hand side would be -(integral of that)

and the right hand side would become F(a)-F(b)

which would simplify to the original equation

so it doesnt really matter weather a>b

[signal canopy]

so the sign of dx doesn't matter on limits?

[shut pagoda]

it does

like you cant just remove the negative

but you can pull it out of the integral and then it would work

[signal canopy]

yes but that negative sign in the image i sent earlier is not pulled out from integral , its already present there

[shut pagoda]

i dont see a -dx

do you mean the range is from infty to r instead of r to infty

[signal canopy]

I was wondering the same why he didn't put (-dx)

thanks i will think about this more

[shut pagoda]

mb i dont really know how to explain this properly

[signal canopy]

noo i don't know much about calculus

My brain needs some time to process it

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[velvet coral]

Thus isn’t the contrapositive also false, and therefore P is false?

[velvet coral]

What are they exactly trying to say here? Isn’t the statement Not P -> C false since we assume Not P is true and C is a contradiction thus it is always false?

[frozen aurora]

C is always false

thus Not C is always true

Not C => P, and since Not C is always true, P is also true

[devout snow]

@velvet coral Has your question been resolved?

[velvet coral]

Yes, but just because Not C is always true does not imply P is true

[frozen aurora]

why not

[velvet coral]

But this is only true “Not C => P” if and only if “not P => C” is true

However the implication “Not P => C” is false

[frozen aurora]

why?

(false => false) is true

[velvet coral]

Since we assumed “Not P” is true and “C” is a contradiction

[velvet coral]

Oh wait wha

I think I don’t understand how not P is false

[frozen aurora]

let's backtrack a bit

[velvet coral]

Ok

[frozen aurora]

we want to show that P is true (goal)

[velvet coral]

Yes

[frozen aurora]

this is equivalent to showing that Not P is false

[velvet coral]

Yes

[frozen aurora]

we deduce Not P => C, where C is false

[velvet coral]

And we do that by assuming Not P is true right

[frozen aurora]

yes

wait

nvm

[frozen aurora]

correct

yes

[velvet coral]

Yeah, but then we arrive at a contradiction which is always false

So true -> false

Which means the implication is false

So then idk how they can use the contrapositive and stuff

[frozen aurora]

yea i am a bit confused too, in our logic class we showed the correctness of a proof by contrdiction very differently

but i think we can get this to work as well

[velvet coral]

Oh would you be willing to share that?

[velvet coral]

Oh yeah okay, we can if you are willing to share what your logic class taught and after

[devout snow]

@velvet coral Has your question been resolved?

[velvet coral]

Not sure why this is exactly true

[frozen aurora]

sorry for the late reply; rereading now, i wouldn't say this is extremely rigorous but the approach is different

instead of using contrapositions we do

$A \wedge \neg B \implies$ absurd

[woven radish]

artemetra

[velvet coral]

Yes, thanks, I think it utilizes the law of excluded middle or smthing

[devout snow]

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[fervent pelican]

My equation for m is different but I think it still passes through a and b just the u has to be changed, is it fine? I don't see it in the marking scheme though and I don't know what the table means jn the marking scheme. Help pls

[fervent pelican]

Talking about part c and b

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@fervent pelican Has your question been resolved?

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[grand ledge]

how would you go and prove this doesnt have a limit?

[grand ledge]

I am not sure I can just say the from 0+ it the same without the floor function, and from 0- is the same but -1

[alpine python]

The square brackets are the floor function?

[grand ledge]

ye

[alpine python]

Yeah your argument works

[grand ledge]

probably shouldve said that

but why

[alpine python]

We can make it more detailed

When x>0, sin^2(x) >0 so the fraction is positive and the floor is >=0

When x<0, sin^2 is again >0 but the fraction is <0 so the floor is <= -1

[grand ledge]

this how you would wrtie it formal?

or is there some formal way to write it

[lusty sapphire]

How "formal" do you want it?

Do you want to find a specific epsilon?

[grand ledge]

no

just something that won't be questions by my lecturer

[lusty sapphire]

calculate the left and right limits

ezpz

[grand ledge]

the 0- is the problem for me

idk how to explain why I am doing the -1

[alpine python]

Because the fraction is <0 because it's a positive over a negative

This isn't enough to say the lim from the left is -1

But it's enough to show the two-sided limit doesn't exist I think

[lusty sapphire]

Show that $\frac{\sin(x^2)}{x}$ would be negatixe for $x<0$ (and some lower bound for $x$ too), and then deduce that $\left\lfloor\frac{\sin(x^2)}{x}\right\rfloor$ would be $\le-1$

[woven radish]

SWR

[grand ledge]

I know this is enough, I just need to explain how I am making the floor function go away with a -1

[alpine python]

Uhh actually there is a hole in my argument

It works for -pi < x< 0

That should fix it

[alpine python]

The floor of x is always <= x

So if the fraction is < 0, the floor of the fraction is the same or even smaller, so it's less than 0 too

You should be able to show floor(x) <= x by looking at the definition of floor

[grand ledge]

I said it like that: For $0^+ : 0 < \sin{x^2} < 1, 0 < x < 1 ==> 0 < \frac{\sin{x^2}}{x} < 1$

[woven radish]

Shachar

[alpine python]

It has the same hole I mentioned before

You can say for x in (0,pi)

Can't say for 0+

[grand ledge]

can I say then aftwer than thar 0+ is in (0,pi)?

[alpine python]

That statement doesn't make much sense to me

[grand ledge]

I do a limit of 0- so I should be able to say something about it

[alpine python]

To be more formal I think you'll need delta-epsilon

Because you need a way to exclude the zeros on the negative x-axis

[grand ledge]

you think if I just write the limit for 0-, than just make it equal to the limit without the floor but -1 they won't question this?

it's a question of is it formal obvious enough that it doesnt need to be proven

[alpine python]

It's not equal

That would be wrong

Or maybe I misunderstood

[grand ledge]

$\lim_{\x\to 0^-} \floor{\frac{\sin{x^2}}{x}} = \lim_{\x\to 0^-} \frac{\sin{x^2}}{x} - 1$

[woven radish]

Shachar
Compile Error! Click the reaction for more information.
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[alpine python]

Uhh

That seems sketchy

[grand ledge]

ye I know

I mean it defintly true

but idk how to get to it

[alpine python]

Has your class taught delta-epsilon?

[grand ledge]

ye

but it seems like a over kill

[alpine python]

Um

[grand ledge]

also never proven something doesnt have a limit using this method

[alpine python]

You have to flip all the quantifiers

[grand ledge]

wym?

[eager nova]

Use taylor

[grand ledge]

didnt learn yet

[alpine python]

Taylor on floor?

[grand ledge]

how would you use delta-epsilon on floor?

[alpine python]

Let epsilon be 1/2

For any delta we can pick x= plus minus delta over 2

Also pick x such that it's not a sqrt of a multiple of pi

Then evaluate at the two x values and one will be >=0 and one will be <=-1 I think

[grand ledge]

idk what you did here

wym by evaluate

isnt delta-epsilon the proof that a limit exists?

[alpine python]

Plug into floor(sin(x^2)/x)

[lusty sapphire]

It can equally prove that a limit doesn't exist

[alpine python]

If you want to go down this road you can start by stating the negation of delta-eps

[grand ledge]

For every $\epsilon > 0$ exists $\delta > 0$ that for every x that $0 < | x - 0 | < \delta$ than $|\floor{\frac{\sin{x^2}}{x}} - L| < \epsilon$?

[woven radish]

Shachar

[alpine python]

Yeah that's the regular version

[grand ledge]

and from here what dhould I do?

[alpine python]

"For all" and "there exists" swap

[grand ledge]

doesnt matter, its not in english

[alpine python]

And you have to negate the implication

[grand ledge]

how?

[alpine python]

It matters

Write the implication as a conjunction

Then negate it

[grand ledge]

so assume it true

[alpine python]

Sorry I meant a disjunction

[grand ledge]

so assume it's not true?

[alpine python]

P=>Q is equivalent to "not P, or Q"

Use demorgans laws

[grand ledge]

we havent learnd those, arent those for logic?

[alpine python]

Yes they are part of logic

[grand ledge]

this is calc

we arent that smart

[alpine python]

I think Spivak's calc book states the negation of delta-epsilon

Maybe yours does too

[grand ledge]

we don't have a book

we just have lectures

and they didnt state this

but im gonna search for it

thanks anyways

[alpine python]

Maybe you can find a way without delta-epsilon

Good luck

You can just say for x close enough, it won't hit a zero of the function

That should be fine

[grand ledge]

isnt this only in math analasys?

[alpine python]

I mean for x close enough to zero, the fraction inside the floor can't be zero

[alpine python]

This is the right idea

But u need <= rather than <

[grand ledge]

but it's not equals

[alpine python]

It can equal 0

When x is sqrt pi

[grand ledge]

if it equals we get to 0/0 and this is another shit I don't want to handle

[alpine python]

No not there

[grand ledge]

I still dont understand why pi

[alpine python]

At sqrt pi

Sin becomes 0

At pi

[grand ledge]

ye but we check for 0+, 0-

how we got to pi

[alpine python]

Yea

[grand ledge]

you mean that sin(x) >= 0 for x in [0, pi]?

[alpine python]

So just state "for x>0 sufficiently close to 0"

Then you can use < and not <=

You can leave as "for 0+" if you want

It's fine

[grand ledge]

For $x \in (0, \pi) [0^+] x >= 0 \and \sin{x^2} >= 0$ therefore $\frac{\sin{x^2}}{x} >= 0$?

is this okay?

[alpine python]

No that's just confusing

[woven radish]

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

[grand ledge]

damn

Im in formal hell

[alpine python]

[alpine python]

You can use this but remove <1 at the end

[grand ledge]

you said I can't just say about 0+

[alpine python]

OK you can say:

For $x\in (0,\sqrt{\pi}),\ 0< \sin (x^2),\ 0<x$ so $0<\frac{\sin (x^2)}{x}$

[woven radish]

Axe

[grand ledge]

then say $0^+ \in (0, \sqrt{\pi})$?

[woven radish]

Shachar

[alpine python]

No just don't say anything about 0+

0+ is not on the number line

It can't be in (0, sqrt pi)

I mean... you can do it if you really want to

It's a little naughty

[grand ledge]

so this can be assumed to just be inside without explicitly telling it is

[alpine python]

Strictly speaking, it's not inside

It's not a number

[grand ledge]

wink

0.000000001

Like this then?

[lusty sapphire]

,rccw

[grand ledge]

magic

[lusty sapphire]

limits of sine functions are always very interesting. This is where a calculus student really starts to question their non-analytic understanding of trig functions

[alpine python]

If this is what you want to say, you will need an upper bound after all. But 1 doesn't work

[grand ledge]

actually the floor function is my problem here, not the sine

[alpine python]

At least not as you wrote it

[olive snow]

Floor ?

[grand ledge]

pi/2 will work?

[lusty sapphire]

btw I would mention $0<\sin (x^2)<1$ for a specific interval of $x$

[woven radish]

SWR

[lusty sapphire]

omg that took too many tries to write

[grand ledge]

we use the square brackets as the floor function in this course

[olive snow]

Ok ok

[lusty sapphire]

That is very unfortunate notation. I don't blame you for that, but it's just very unfortunate.

[alpine python]

Yeah you can put < 1 back

But...

[grand ledge]

why does it so hard to prove something so obvious 😦

[alpine python]

We haven't proven sin(x^2)/x < 1

On (0, sqrt pi)

[grand ledge]

pretty ez tho

sin <= 1

oh nvm

idk what sin(x) equals for every x

I hate my life

[alpine python]

Maybe you can argue the leg is shorter than the arc on the unit circle

Or

Just assume it

[grand ledge]

when tf did we moved to trigo

[alpine python]

[grand ledge]

aint no way they meant for me to go this deep into this

[alpine python]

Then just assert sin(x^2)/x < 1

On that interval

[lusty sapphire]

I think I have an idea

if you're still feeling up to working on this

[grand ledge]

for sure if it will solve this

[wise storm]

the way I sorta "see" it but depends on having seen the sinc function before is:

lim x->0 of sin(x)/x is 1 so

floor(sin(x^2)/x) = floor(x * sin(x^2)/x^2) is approximately floor(x) near x=0 which has no limit as x->0 either

[lusty sapphire]

you can take a few minutes to mentally reset though if you want. I'm still writing out my idea to minimize any possible confusion as I attempt to explain it

[brittle inlet]

You sometimes have to think creatively, you know

[grand ledge]

there is no problem of creativity just of formality

[grand ledge]

thanks lol

@wise storm i tought about it, and it might even be possible to solve it like this, it's just invloves saying for 0+ it's continous in (0, 1) and for 0- it's continous in (-1, 0) than can just put the lim inside the floor. but again idk how to formally write this

[alpine python]

the problem you're having at the moment is more to do with the sin than the floor

[wise storm]

yeah I should have read more of the thread here but it was kind of long.I think what you're suggesting here is perfect so nothing I can really add I think haha

[alpine python]

you can pull the limit inside continuous functions BUT

i'm worried in this case for two reasons

the limit is at the endpoint of the interval of continuity, and

[lusty sapphire]

floor is not continuous at x=0

[alpine python]

right

and we're considering the interval (0, sqrt(pi))

we need 0 < sin(x^2)/x < 1 so that the argument of floor is within (0,1) where it's continuous

i think

[grand ledge]

I wrote

For $x \in (0, \sqrt{\pi}), 0 < \sin{x^2} < 1, x > 0$ so $0 < \frac{\sin{x^2}}{x} < 1$

[woven radish]

Shachar

[alpine python]

the conclusion that sin(x^2)/x < 1 doesn't follow from the premises

[grand ledge]

so I just need to find what will make it follow

[alpine python]

regarding pulling the limit inside floor

i found this problem:

Prove that if $f$ is continuous at $l$ and $\lim_{x\to a} g(x)=l$, then $\lim_{x\to a} f(g(x)) = f(l)$.

[woven radish]

Axe

[alpine python]

in our case lim(g)=1 and f isn't continuous at 1

so it doesn't apply

[devout snow]

@grand ledge Has your question been resolved?

[grand ledge]

what idea btw?

[lusty sapphire]

Just finished writing it out.

Okay we can go through it'

[alpine python]

wait lim(g)=0

[lusty sapphire]

I'm pretty confident I can help you get this done @grand ledge

[grand ledge]

I believe

[lusty sapphire]

Does this much make sense right now?

[grand ledge]

yup

[lusty sapphire]

Okay, and here will be the general approach. Let me know if you have questions:

[grand ledge]

you don't need to prove it?

[lusty sapphire]

We're going to prove it. I'm just giving you the outline of the overall proof

[grand ledge]

ok

understood

[lusty sapphire]

I'm sure you know what the floor function is, but I want to express this point, as it will be how we get this all done

[grand ledge]

yes

[lusty sapphire]

And do you see how this would get us to show that the limit does not exist?

[grand ledge]

ok, that makes sense

[lusty sapphire]

And does this part make sense in how we can prove the above?

[alpine python]

you changed the denominator

[lusty sapphire]

yup

what about it?

oops I forgot that bit

[lusty sapphire]

I meant to mention this first

[lusty sapphire]

and then this

[lusty sapphire]

@grand ledge does the logic here make sense?

[grand ledge]

ye

a bit confused

with the all theta thing

but I guess it's just to say that sin(theta) between 0 and 1 here?

[lusty sapphire]

I was saying just that $\sin(\theta)>0$. The upper bound $1$ will not be important for us, but there is no harm in including it.

[woven radish]

SWR

[grand ledge]

ok

you can continue then

[lusty sapphire]

So now, finally, we get to where the bulk of your confusion has been

We will need to show this bit geometrically

If we can show this, do you see how the rest of the proof would fall in line now?

[alpine python]

ye the leg is shorter than the arc