help-42

[leaden thunder]

it depends on the x value

between 1 <x < 10, the orange is on top, then that means that function is greater than the blue one

at x=10 it switches

[untold tapir]

so its orange? cus the lower limit is 1 and upper is 10 and we dont count when it switches

[leaden thunder]

what is "its"

[leaden thunder]

are you asking if this sentence is correct?

[untold tapir]

bro i meant upper function the equation is intergal( upper function - lower function) for area between curves

idk which one is the negative

[leaden thunder]

the upper function is the top one

so it is orange yes

[untold tapir]

ok ty

[calm coral]

@untold tapir Has your question been resolved?

[calm coral]

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[balmy sentinel]

I need to find n so that En is a natural number. The possible answers are in the thickened parenthesis. I got this far but I'm not sure what to do from here

[cobalt briar]

you can write the numerator as ∑k(k+2)

[balmy sentinel]

hmm, and after that?

[cobalt briar]

∑k^2 = n(n+1)(2n+1)/6
2∑k = n(n+1)

so you get

n(n+1)(2n+7)/6

numerator simplifies to n(n+1)(2n+7)

demnominator is n(n+1)(n+2)

so you get 2n+7/n+2

now you can put values of n

[balmy sentinel]

hmm let me give it a try

[cobalt briar]

sure

[balmy sentinel]

is it okay like this?

[cobalt briar]

yeah

[balmy sentinel]

woooo! Thank you once more

[cobalt briar]

no worries

[balmy sentinel]

.close

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[balmy sentinel]

no ideas once more

[nocturne heron]

what’s the sign of a_n

[mortal orbit]

maybe try to upper bound the integrand

[cobalt briar]

yeah

[calm coral]

@balmy sentinel Has your question been resolved?

[balmy sentinel]

like this?

[mortal orbit]

uh

sin is not non-negative

[balmy sentinel]

uhh right

then I might just no know how to do what you suggested

what does upper bounding an integral mean again

waiiit

if i start with sin(nx)

[mortal orbit]

you're upper bounding the wrong quantity

[cobalt briar]

yeah

[balmy sentinel]

ohhhh

okay i got it

[mortal orbit]

that's more like it

[balmy sentinel]

like this right?

[cobalt briar]

yeah correct

[calm coral]

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[calm coral]

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[cold basin]

Hello, im getting cosphi=sqrt(3)/4

Given cube ABCDA1B1C1D1 where ABCD is the base. Given point M midpoijt of BB1, point N midpoint of A1B1 and point K, midpoijt B1C1, find the cosine of the angle between the planes (ABCD) and (MNK)

I see this is angle HMD, where H is the midpoint of NK

[glad parrot]

Open a new one

[cold basin]

Why

Oops

[glad parrot]

Good question

[calm coral]

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[still terrace]

@eternal shard hi

[still terrace]

for b)

do i just have to put 0,5 as skalar

for AB

[calm coral]

@still terrace Has your question been resolved?

[short salmon]

german 0_o

i cant read

[calm coral]

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[glad parrot]

Asking for AB midpoint coordinates

[calm coral]

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[green cosmos]

what would i use as a comparison for problem 20?

[keen grail]

For n > 3, n^n < n^2

Or

Rather the reciprocal I mean

[eternal shard]

write some terms down

then it might tell you immediately how you might estimate it

[green cosmos]

1/n^1

i mean

1

1

[eternal shard]

1/1^1

[green cosmos]

then

1/4

[eternal shard]

1/2^2

1/3^3

1/4^4

1/1^1 + 1/2^2 + 1/3^3 + 1/4^4 + ...

if you compare it term wise

[green cosmos]

i see

i get it now

so 1/n^2

[eternal shard]

you can always choose 1/2^k

[green cosmos]

https://tenor.com/view/cat-smiling-cat-grey-cat-cute-cat-gif-2949633679466643461

[eternal shard]

and argue with the geometric series

[green cosmos]

for 21 i can use 1/n right

as a comparison

hm

1/sqrt1+1 = 1/sqrt2

i looked at these

https://www.youtube.com/watch?v=LBxYQ0TJxYM&ab_channel=TheOrganicChemistryTutor
https://www.youtube.com/watch?v=oZtAgihok5s&t=55s

I watched through both of these

uhm

[eternal shard]

You just need to make 1/sqrt(n^2+1) smaller by making the denominator bigger

[green cosmos]

if the big series converges then the smaller one should converge as well

if small series diverges then big seriesi diverges

bigger

[eternal shard]

hey

[green cosmos]

oh

[eternal shard]

how can you make the denominator bigger

[green cosmos]

its add to denominator

smaller

yea

[eternal shard]

what's bigger than n^2+1 for sufficient n

[green cosmos]

i dont know what does the sufficient n part mean

[eternal shard]

(the goal is to rid the +1)

[eternal shard]

it does mean that we only care that the inequality works not for a specific n

like

you could have an inequality

that starts working if n = 10000 and bigger

or n = 4

the point is you just have to assure that such n exists

your series have infinite terms which mean n goes to infinity

so when using comparison test, you need to find inequalities that certainly work out for large values of n

[short salmon]

for all but finitely many n!

[eternal shard]

most of the times removing the constant and adding a factor does the trick

like

n+100 <= 2n for example for some n at some point will be true (or large amount of values of n)

[green cosmos]

so

2n^2

[green cosmos]

pikiop

[eternal shard]

yea or nicer 4n^2 due to the square root

now try to show why

it works

why 1/sqrt(n^2+1) >= 1/sqrt(4n^2)

[green cosmos]

i can see it being 2n^2

not 4n^2

[green cosmos]

how does the 4n^2 happen due to sqaure root again?

[eternal shard]

2n^2 works too

i was just saying 4n^2 is more nicer

[green cosmos]

icic

[eternal shard]

both work but i still wanna see if you can show it

[green cosmos]

im just curious

am i able to do 1/sqrt(n^2+1) <= 1/sqrt(n)

[eternal shard]

what do you think

[green cosmos]

yes

[eternal shard]

prove it

[green cosmos]

ok

[eternal shard]

try to derive a true statement

1/sqrt(n^2+1) <= 1/sqrt(n)
since both sides are positive what could you do

[green cosmos]

i just know that 1 become negilible in the statement

and

i just know the behavior of 1/sqrtn when compared to the original function

can i get a example

[eternal shard]

example of what?

[green cosmos]

of a true statement being shown, likw how id prove it

[leaden thunder]

you should try to prove adonis' inequality

do easy problems to buid skill to do harder ones

[green cosmos]

prove that this is true?

one has n^2 +1 on the bottom while the other one just has n

making it smaller

[leaden thunder]

that's not really enough to give a "proof"

[green cosmos]

uhh

[leaden thunder]

and he probably meant 1/sqrt(n^2) instead of 1/sqrt(n)

[eternal shard]

no

[leaden thunder]

oh i see

[leaden thunder]

he was responding to this

[green cosmos]

1/sqrt(n) is a p-series you can compare to because p > 1 making it divergent comparing it to a larger thing 1/sqrt(n^2+1)

[eternal shard]

she asked that lol

[leaden thunder]

yea i missed it

[eternal shard]

so i asked her to prove it

for her own good skull

[eternal shard]

1/sqrt(n) is larger than 1/sqrt(n^2+1) so that doesnt show it diverges

you have to find a smaller series

dct: show it converges by finding a greater convergent series
show it diverges by finding a smaller divergent series

like

sum f(n) <= sum g(n) where the sum of g(n) diverges, but sum of f(n) might just be small enough converging maybe

so you have to find instead a smaller diverging series

[calm coral]

@green cosmos Has your question been resolved?

[green cosmos]

this requires the limit comparison right?

because 1/sqrt(4n^2) cannot be proved by the direct comparison ytest

like it doenst fit into the rules

[green cosmos]

show it divierges by finding a smaller divergent series while 1/2n is greater

[native fox]

what question are we discussing?

[upper sparrow]

are you sure?

[green cosmos]

i keep thinking about the question

[green cosmos]

prove that this works

how do i prove it?

[green cosmos]

uhh

yes

because 1/2n

is greater

you can only find a greater convergent series

dct: show it converges by finding a greater convergent series
show it diverges by finding a smaller divergent series

you find a divergence by a smaller divergent series

1/2n is divergent

[upper sparrow]

(well, the fact that n >= 1 should give you that n^2 >= 1, thence "larger denominator smaller fraction")

[copper epoch]

All u have to prove is
n^2+1 <= 4n^2

And u are done

[green cosmos]

isnt it n^2+1 <= 2n^2

2n

i mean

[copper epoch]

I mean u reverse it so the < changes direction then whatever is inside that sqrt can be compared

[upper sparrow]

RHS is in turn <= 4n^2

[green cosmos]

yea

okay

[upper sparrow]

You could equally work with getting sqrt{2} * n and all the subsequent steps, which get you the same conclusion

[green cosmos]

.close

[calm coral]

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[calm coral]

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[inner urchin]

simple algebra equation

[latent steeple]

What have you tried

[inner urchin]

im ot sure how to answer the question

I tried y2-y1/x2-x1

I need help thorfinn

[latent steeple]

Yeah what did you get

[inner urchin]

-12.5/18.5

[latent steeple]

Just do (25-0)/(0-35) to find gradient. Simple

[inner urchin]

how did you get 25-0

ohh

both intercepts

[latent steeple]

Oops my bad -25-0

I took y2=-25 and y1=0

[inner urchin]

im saying the x and the y intercept

[latent steeple]

So did you get the gradient

[inner urchin]

-5/-7

5/7

then u multiply by 7

.close

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[clear stone]

Hi so like I figured out my answer for the angle of this question, but I'm debating whether the general equation should be πn or 2πn since one revolution of tan is π, I'm unsure though.

[clear delta]

,tex .unit circle

[potent lotus]

hayley (now with real hay!)

[clear delta]

you can use this to identify all the spots where tanx = √3 or -√3

and then add 2πk

[clear stone]

Right so it is 2πk

Thank you so much

.close

[swift laurel]

you can either leave it like that and use 2pi n, or notice that if you used pi n instead then solutions (1) and (3) become the same (and similarly with solutions 2 and 4)

[calm coral]

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[silver tapir]

isnt tan π periodic

[swift laurel]

yes

[calm coral]

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[simple bridge]

Need help with this boolean algebra problem

this is the whole question

I think the expression for the network in figure 7 is WXY(Z+Z'(S'+V))

but I'm having trouble showing the second expression is equivalent to the first one I came up with

the closest I can get applying the properties is WXY(Z'(S'+V)+Z(SW+X))

am I doing something wrong or is there something wrong with the question

<@&286206848099549185>

[calm coral]

@simple bridge Has your question been resolved?

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[lean aspen]

How do I reflect this graph across the Y axis?

[lean aspen]

I want to reflect the blue graph across the y axis

[clear delta]

replace all instances of x with (-x)

[lean aspen]

that just gave me the green graph

actually not quite

I see

this is working

gotcha

ight now how do I rotate it?

er

how did I even graduate high school without understanding graph transformations

maybe I'm not ready to rotate it yet

[calm coral]

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[unborn plover]

ok

[unborn plover]

How do I reflect this graph across the Y axis?

I want to reflect the blue graph across the y axis

[brazen elbow]

negate every x

[unborn plover]

that just gave me the green graph
actually not quite
I see
this is working
gotcha
ight now how do I rotate it?
er

[brazen elbow]

what you did was only negating everything

[void umbra]

Sub all x with (-x)

[brazen elbow]

more specifically, if you want to reflect f(x) through the y-axis, then you need to plot f(-x). your mistake is instead plotting -f(x)

[unborn plover]

ok

[calm coral]

@unborn plover Has your question been resolved?

[calm coral]

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[zealous loom]

Look at #14
The book says that the answer to this example involves (m-n), but how does M suddenly become positive?

[jovial pumice]

,rotate

[jovial pumice]

does the book say (m-n)(X-Y)^3?

[zealous loom]

Yes

[jovial pumice]

i dont think u have done anything wrong

[zealous loom]

Then why did m become positive and n become negative?

In the answer

[jovial pumice]

see u wrote m(X-Y)^3 as -m(Y-X)^3

but in the book method they did, n(Y-X)^3 as -n(X-Y)^3

it doesnt change the answer, just diff methods

urs is correct too

[zealous loom]

Doesn't it?
(-m) + n ≠ (m-n), because
(-10) + 5 = -5, but 10 - 5 = 5

[jovial pumice]

(m-n)(Y-X)^3 = -(m - n)(X-Y)^3, does this make sense

[zealous loom]

Nope

Because I'd do the answer as (y-x)³ × ((-m)+n)

[jovial pumice]

u have literally wrote the same thing 😭

[jovial pumice]

there is prolly some miscommunication

[zealous loom]

Have I? The - is in a different place (kinda) and there's a plus between m and n

[jovial pumice]

does (X-Y)^3 = -(Y-X)^3 make sense?

$(X-Y)^3 = -(Y-X)^3$

[potent lotus]

Piyush

[zealous loom]

Yes

[jovial pumice]

ok thats good

so if we multiply (m-n) on both sides we would get --

$(m-n)(X-Y)^3 = -(m-n)(Y-X)^3$

[potent lotus]

Piyush

[jovial pumice]

$now, (m-n) = -(n-m)$

[potent lotus]

Piyush

[jovial pumice]

$so, (m-n)(X-Y)^3 = (n-m)(Y-X)^3$

[potent lotus]

Piyush

[zealous loom]

Huh

[jovial pumice]

😭 we are just taking -1 common

i dont know how else to explain😭

[zealous loom]

I see what you mean

But

[zealous loom]

The answer in my book doesn't switch m and n places

[jovial pumice]

oh?

[zealous loom]

The book says (m-n)
You say (n-m)

[jovial pumice]

see the left hand side😭

[zealous loom]

Right

I'm an idiot

[jovial pumice]

no no, its ok

[zealous loom]

I forgot my x-y is also on the other places

[jovial pumice]

😭

[zealous loom]

So if we were to rotate them both, it's correct

[jovial pumice]

yess😭

[zealous loom]

Thank you a lot

You're awesome

[jovial pumice]

ur wlcm

have a great day

[zealous loom]

.close

[calm coral]

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[zealous loom]

You too

[calm coral]

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[fluid swallow]

Help, I am a bit sick today … so please be patient with my responses I might be very distracted…

[fluid swallow]

Let ( P ) denote the ( \mathbb{R} )-vector space consisting of all polynomials over ( \mathbb{R} ) in the variable ( x ).
The ( \mathbb{R} )-linear transformation ( T: P \to P ) is defined as ( T(f) := \int_x^{x+1} f(t) , dt ).

\begin{itemize}
\item[(a)] Show that if ( f \in P ) has degree ( m ), then ( T(f) ) also has degree ( m ).
\item[(b)] Prove that ( T ) is injective.
\item[(c)] With ( P_n \subset P ) the ( \mathbb{R} )-subspace consisting of the polynomials of degree ( \leq n ), show that
as a map from ( P_n ) to ( P_n ) the transformation ( T ) is surjective.
\item[(d)] Prove that ( T: P \to P ) is bijective.
\end{itemize}

[potent lotus]

Emmaaaaa

[fluid swallow]

Question statement

(a)

Proposition for every natural number $m$ we have the following statement hold, $f\in \mathcal{P}$ is a polynomial of degree $m$ such that $Tf$ has a degree of $m$ too.

Base step: $m=0$ then $f$ is given as a constant polynomial of degree $0$, let $c \in \mathbb{F}$ be some scalar such that $f(t)=c$ thus $T(f)=\int_x^{x+1}c\text{ dt}=c(x+1)-cx=c$. This verifies the base step.

Inductive hypothesis: for some natural number $m\in \mathbb{N}$, and $m>0$ we have the following statement: $\deg{p}_m=\deg T(p_m)$ where $p_m\in\mathcal{P}$ is a polynomial of degree $m$

Then for $p_{m+1}$ and some scalar $c\in \mathbb{F}$ we have 
\begin{align*}
\deg{T(p_{m+1})}=&\deg\bigg(\int^{x+1}_x ct^{m+1}\text{dt}\bigg)\\=&\deg \bigg[t^{m+2}\bigg]^{x+1}_x\\
=&\deg\bigg[t^{m+2}\bigg]^{x+1}_x\\=&\deg\bigg(\sum^{m+2}_{j=0}\begin{pmatrix}
    m+2\\j
\end{pmatrix} x^{m+2-j}-x^{m+2}\bigg)\\
=&\deg\bigg(\begin{pmatrix}
    m+1\\1
\end{pmatrix}x^{m+1}+\begin{pmatrix}
    m+1\\2
\end{pmatrix}x^m+...\bigg)\\=&m+1
\end{align*}

It is clear that our induction implies the inductive hypothesis thus the transformation $T$ indeed preserves the degree of polynomial $p$

(b) Let $p$ be any polynomial then $T(p)=P(x+1)-P(x)$ where $P(x+1)$ is a horizontal shits of the polynomial $P$ that said by the polynomial identity, the sum $P(x+1)-P(x)=0$ if and only if $P$ is the zero polynomial. hence we can conclude conclude that $\ker T={0}$, immediately it follows that $T$ is an injection.

[potent lotus]

Emmaaaaa

[fluid swallow]

(c) By the dimension theorem:
\begin{equation}
\dim \text{dom }T=\dim \text{im} T +\dim \ker {T}=n<\infty
\end{equation}

By our conclusion of $(b)$ it shows that $\ker T={0}$, the injectivity ensures the surjectivity of $T$ which implies that $T\in \text{bij}(\mathcal{P}_n)$

(d) Observe the transformation $T$, for some arbitrary scalar $c\in \mathbb{F}$ and $p, q \in \mathcal{P}$ we have:
\begin{align*}
T(cP+q)=&\int^{x+1}{x}(cp+q)(t)\text{dt}\=&c\int^{x+1}{x}p(t)+\int^{x+1}_{x}q(t)\=&cT(p)+T(q)
\end{align*}

Which shows that $T$ is linear. Furthermore $\forall{p}\in P_{\deg{p}}\subset\mathcal{P}$ we know that $T:P_{\deg{p}}\to P_{\deg{p}}$ is surjective, hence $T\in \text{Bij}(\mathcal{P},\mathcal{P})$

Formalization: $V_n$ are some vector spaces such that we have $\mathcal{P}=\bigcup\limits^\infty_{i=1}V_n$

Assume the induction for $(a)$ for finite case, we claim further that the mapping $T:V_n\to T(V_n)$ is an isomorphic mapping preserving the degree polynomial, that is, in particular for every $p\in V_{n+1}$ there exists some $q\in V_n$ such that for some scalar $c\in \mathbb{F}$ we have $p(x)=q(x)+cx^{n+1}$.

Note that \begin{align*}
\deg{T(p)}=&\deg{T(q)+cT(x^{n+1})}\
\leq& \deg T(x^{n+1})\=&\deg\bigg(\frac{1}{n+2}\bigg(\begin{pmatrix}
n+2\0
\end{pmatrix}x^{n+2}-x^{n+2}\bigg)+\begin{pmatrix}
n+2\1
\end{pmatrix}x^{n+1}+\begin{pmatrix}
n+2\2
\end{pmatrix}x^n+...\bigg)\=&n+1
\end{align*}
Showing that $T$ indeed preserves degree hence an isomorphic mapping. Hence bijection

[potent lotus]

Emmaaaaa

[fluid swallow]

I actually don’t know if it’s necessary to use transfinite induction nor did I know how to use it properly but I felt it wasn’t sufficient to conclude for d so I added the formalization part in which I don’t really know if it’s needed

[mortal orbit]

uh

I have to review everything properly

so first of all

this equality is not well argued

it might be true, but the way you wrote it there's no way to verify

then

This I'm not sure with your reasoning

first of all P could be any constant polynomial

and even then you have to prove it in more detail

c) is alright

d) is all over the place, the "formalization" is messier than the original argument

it's alright to reprove that T is linear, but in the same time:

• 1. it's already assumed that T is linear in the question statement
• 2. if you did want to prove that T is linear by yourself, you should have proved it earlier, in question (a) or (b), when you explicitly used linearity to show the results of those questions

so this part is misplaced or removable entirely

A little more rigorous proof to show that T is surjective is enough for (d)

such as: $\$
"Let $p \in \mathcal P$. If $p = 0$, then $T(0) = 0$. Otherwise, we let $n = deg(p)$. since $T:\mathcal P_n \to \mathcal P_n$ is surjective, there exists $q\in \mathcal P_n \subset \mathcal P$ such that $T(q) = p$. Therefore $T:\mathcal P \to \mathcal P$ is surjective."

[potent lotus]

rafilou is not not born in 2003

[calm coral]

@fluid swallow Has your question been resolved?

[fluid swallow]

yes I actually don’t really know if I can prove the infinite dimension case so I added the formalization part… let me read line by line

I am sorry I am bit sluggish I had a fever

[mortal orbit]

it's alr

[fluid swallow]

I didn’t read properly today I got so rushed 😭😭😭

[fluid swallow]

Should I expand the integral and then argue the degree?

Let me see if I can fix tonight or else I fix it tomorrow

[mortal orbit]

no I mean for question a), this is where the induction hypothesis comes into play

your polynomial of degree m+1 isn't just cx^(m+1)

it's cx^(m+1) + q(x) where deg(q) <= m

so

[fluid swallow]

That’s how you linked to it, I was thinking about it like I should use hypothesis

Let me fix it now

[mortal orbit]

deg(T(p)) = deg(T(cx^(m+1)) + T(q))

keep that T(q) in the corner while you do the calculations you did before

and finally when you get around that line

[fluid swallow]

Yes, that sounds much better… 🥰🥰🥰

[fluid swallow]

You’ve always been a hero 🥰🥰

[mortal orbit]

the T(q) and the other x^m, x^(m-1)... terms

are negligible for the degree

and so the degree is m+1

by the way, I'm just noticing but in your computations in (a)

you reasonably get rid of constant coefficients

because they're in factor of the whole polynomial

but now that you have T(q) aside

mmh

Ok I have an alternative to solve those problems

just compute deg(T(cx^(m+1))) in the meantime

(so we're not computing deg(T(p_(m+1))) yet)

so now no need to amend the rest of the computations

and after that

just say deg(T(p_(m+1))) = deg(T(cx^(m+1)) + T(q))

T(cx^(m+1)) has degree m+1

T(q) has degree at most m

you have your result

[fluid swallow]

That’s clever I didn’t think of it 🥰🥰🥰

[mortal orbit]

yeah so linearity is already at the heart of question (a)

and that's also why we're doing a proof by induction

[fluid swallow]

Yes, i feel like my argument has been so convoluted in many way…. I will take a screen shot and sleep for now I am a little bit too headachey I will send you messages if that’s fine in the morning 😭😭

[mortal orbit]

if the point of the proof (induction hypothesis) isn't used in your proof, start questioning why you're using it in the first place

[fluid swallow]

True, I was thinking of using it but I just felt I didn’t know how it is related

Is it fine if I send you message in a couple of hours, I don’t know if you’ll be busy by then 😭😭

[mortal orbit]

uh it's alright

[calm coral]

@fluid swallow Has your question been resolved?

[calm coral]

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[mild hull]

I quite literally have no idea where to start on this
Ps: the chinese wordings translates into "Based on this diagram" so it isnt important

[vital blade]

form a triangle

[mild hull]

How form

[vital blade]

do u understand

[mild hull]

Yepp im solving it rn

Yes i got it tyvm, never thought id be so simple lol

.close

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[echo birch]

I'm trying to compute the integral $$I = \int_{e^{-\pi/2}}^{e^{\pi/2}} (\sin^2(\ln x) + \sin(\ln(x^2)))dx$$. I figured I'd just write the sines in term of Euler's formula, get a super easy integral with the logarithms cancelling out, and then just do it. Apparently, that doesn't work.
Here's what I did:
$$
I = \text{Im}\left[\int_{e^{-\pi/2}}^{e^{\pi/2}} e^{2i\ln x} + e^{2i \ln x} dx\right]
$$
$$
I = \text{Im}\left[2e^{2i}\int_{e^{-\pi/2}}^{e^{\pi/2}} x dx\right]
$$
$$
I = \text{Im}\left[e^{2i}(e^{\pi} - e^{-\pi})\right] = \sin(2)\left[e^{\pi} - e^{-\pi}\right]
$$
The correct answer, however, is
$$
I = e^{\pi/2} - e^{-\pi/2}
$$
I'm very confused about where I messed up, because it looks alright to me.

[echo birch]

nooooo

im not done yet

sorry

[potent lotus]

avipi

[echo birch]

ok im done

[manic hawk]

It looks like you should have gotten pi/2 in your exponents when you evaluated the integral

[echo birch]

I don't understand.. the integral evaluates to x^2/2, the 2 next to the exponential cancels it out, and the squaring removes the 2 in denominators of the exponents, right?

[manic hawk]

Oh,I overlooked the x, sorry. One minute

But I do see the other mistake. It's your handling of sin^2

[echo birch]

oh

[manic hawk]

You need to use a trig rule on that first

Imaginary part and squaring don't commute

[echo birch]

oh

1-cos(2x)/2 is fine, right?

[manic hawk]

It looked fishy to me because the two terms in the integrand are different but you evaluated them to be the same

[echo birch]

hmm

got it

then i'd need to break it up into 2 integrals, one real part and one imaginary

eh

thanks!

i think i have it now

.close

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[remote mural]

Is this some known binomial identity?

[rocky tendon]

Yeah

[
\sum_{i=0}^{d} \binom{2d - i}{d - i} = \binom{2d + 1}{d}.
]

Explanation:
By substituting (k = d - i), the sum transforms into (\sum_{k=0}^{d} \binom{d + k}{k}). This matches the identity:
[
\sum_{k=0}^{n} \binom{n + k}{k} = \binom{2n + 1}{n},
]
which holds for non-negative integers (n).

[potent lotus]

Boas

[calm coral]

@remote mural Has your question been resolved?

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[eternal shard]

hi

[eternal shard]

I think I can cook up a proof

[ancient thistle]

what's your definition of e^z

[eternal shard]

If $z = x+iy$ then $e^z = e^x \cdot (\cos y+i \cdot \sin y).$

[eternal shard]

x \in \R and y \in (-\pi,\pi]

I tried to use the Ln(w) := ln(w)+iarg(w)

[ancient thistle]

okay then isn't this kinda just apparent

unless you want to prove everything from scratch

cos(y) + isin(y) is surjective into the unit circle

and e^x is surjective onto R_>0

[eternal shard]

yes

but is that enough of an argument

[ancient thistle]

why wouldn't it be

[eternal shard]

like sometimes, there are ppl who say you need to use the definition in order to be mathematically correct

[eternal shard]

and i was thinking kinda isnt that just stating things

[ancient thistle]

well e^x cos and sin are defined in real analysis

you should have their properties available to you

that's why i said unless you want to prove everything from scratch

[eternal shard]

ok i was misunderstanding you mb

[ancient thistle]

in which case you have to dig up whatever definitions of e^x cos and sin you are using

and then that's like a whole different can of worms

[eternal shard]

i see

i am just anxious

like i was thinking what if i was asked to prove something like that

am i allowed to use facts

[ancient thistle]

then you assume the well known properties of e^x cos and sin

the real versions not the complex versions

[eternal shard]

oh ok i see

i see see

ok well then it's obvious

can i still show you what i had in mind

[ancient thistle]

what did you have in mind though

you're not gonna prove properties of e^x cos and sin on the spot

that's like a whole slew of lemmas from real analysis

[eternal shard]

it will be probably dumb, but i guess thats how i learn

[ancient thistle]

i mean this is tautological

the fact that you have log|w| is under the assumption that e^x has an inverse

which already presupposes that it is surjective

similarly for arg(w)

[eternal shard]

ok i needed this clarification, and by tautological you mean that it's meaningless true?

[ancient thistle]

to even begin writing those things down, you need that e^x is surjective onto R_>0, and that cos(y) + isin(y) is surjective onto the unit circle

otherwise the expressions log|w| and arg(w) are meaningless

or at the very least their properties are meaningless

[eternal shard]

ok makes sense

sparked my brain

thank you so much!

[ancient thistle]

[eternal shard]

.solved

[calm coral]

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[orchid palm]

Anybody here know how to do 1's and 2's complement

[orchid palm]

I have some questions because something is bugging me

I'll post my work in a second

isn't that the end step?

It mentions that in the case of positive integers, the remaining bits are identical to the binary expansion of the integer

well if that's the case, then "1's complement" for a positive integer isn't really taking the complement of anything, just writing it as is in binary

or am I reading it wrong?

[velvet osprey]

mm not quite no

complement means the bit flipping thing

like specifically the way negative integers are stored

btw you might not wanna draw your long division symbol as a root sign

[velvet osprey]

positive integers are stored in the same way whether you're going 1's comp or 2's comp

[orchid palm]

hmmmm

ok

so then what would it be

?

I flipped the bits

but that doesn't make sense anymore

like at all

all I did was make 22 (positive) into -23

that's what one's complement does on a positive* value?

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[leaden thunder]

<@&268886789983436800> spam

[calm coral]

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[calm coral]

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[gilded pendant]

Hi, good day! I just wanna know what epilson delta is.. and can't I just solve this normally lol

[leaden thunder]

you can solve it normally yea

but if the instructions tell you to use eps-delta, then you should probably use eps-delta

[calm coral]

@gilded pendant Has your question been resolved?

[gilded pendant]

what's eps-delta?

[leaden thunder]

eps is a common abbreviation for epsilon

[tulip spear]

have you tried le google?

le wikipedia?

le (insert any relevant online source that will provide you an answer within a few minutes)?

(https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/01%3A_Limits/1.02%3A_Epsilon-Delta_Definition_of_a_Limit)[Epsilon-Delta Definition of a Limit]

Are you

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

how does it work

[gilded pendant]

I dont understand em

[tulip spear]

see my rant above

[gilded pendant]

it says page not found tho

nvnm

nvm

[tulip spear]

let me re-link the link

nevermind

discord and its stupid hyperlink

[gilded pendant]

it's fine now

thank you!

um...

this is ivt

I'm unfamiliar with the m and the f(c)...

is the f(c) just the x?

<@&286206848099549185>

also is m jusy a given?

[calm coral]

@gilded pendant Has your question been resolved?

[balmy grove]

Usually the notation of f(x) allows x to be substituted for any other variable "name" like c in the inner example. This one demonstrates finding the coefficient c from the result of m.

I think this m refers to the value possible for all X, and value possible for all Y.

Or actually, the function results.

f(a), f(b) ... f(y), f(z), the parameter name is declared as the one within the parenthesis.

f(x) and g(x) refer to different functions, their names are f and g.

It's all about the notation.

[calm coral]

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[astral mural]

Hello

[astral mural]

Can someone help me understand this DE?

Not sure where 5 and 0 and Q(t)/10 is coming from

[swift laurel]

it comes from the rather verbose line above.
flow rate of liquid entering = 5 L/min
concentration of substance in liquid entering = 0 kg/L (fresh water)
flow rate of liquid exiting: 5 L/min (same rate as entering)
concentration of substance (salt) in liquid exiting: current concentration of salt in the tank = (amount of salt)/(volume of tank) = Q(t) / 10L

[astral mural]

I'm confused where 0kg/L is mentioned

[swift laurel]

fresh water means no salt

[astral mural]

Ohhh it's just the rate of 5L/min coming in

Also, how did they come up with the equation for dQ/dt?

[swift laurel]

(rate of change of amount of substance in the tank) = (rate of substance entering the tank) - (rate of substance exiting the tank)

(rate of substance entering the tank) = (flow rate of liquid entering the tank) * (concentration of substance in liquid entering)

(rate of substance exiting the tank) = (flow rate of liquid exiting the tank) * (concentration of substance in liquid exiting)

[astral mural]

I see

Could you also explain the part about where t = 0?

I don't understand how they decided to use Q(t) > 0 and subbing that as Q(0) = 4

[swift laurel]

Q(t) is an amount of substance, it can't possibly be negative

and Q(0) is the initial amount of salt in the tank, which is 4 kg

[astral mural]

How do we know that Q(0) is 4?

Ohh

What about where C = ln(4)?

[swift laurel]

that is from plugging in the initial condition to the equation

[astral mural]

Kind of confused why they are subbing numbers for a function

[swift laurel]

the equation has to be true for all t, so if you substitute in the initial condition that allows you to solve for the integration constants

[astral mural]

So Q(0) = 4 and t = 0?

[swift laurel]

yes, substituting that in enables you to solve for C

[astral mural]

Ohhh that makes so much sense

Thank you so muchh

I was struggling very

.close

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[tacit socket]

hello people?

i belive there are math scholars here

[winter elbow]

We have scholar mate instead

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[wind vault]

Can someone help me understand the approach my teacher made in solving this question.

[wind vault]

I assumed you were to find the area between y=3x^2 and y=4-4x, then subtract that area by the one between 8x^2 and 4-4x

But the answer took a very different approach that I hardly understand

[short salmon]

just to be clear, this is #2, right?

[calm coral]

@wind vault Has your question been resolved?

[wind vault]

Yes

[short salmon]

I don't really have the time to look through all the work carefully at the moment, but the general approach is to split the integral based on the x values where curves touch, and subtract integrals of the lower function from the higher function (since you're working in non-negatives)

[wind vault]

Ohhh, that makes so much more sense

Tyyy

[short salmon]

oh, I'm really glad that my answer was able to help despite not really looking at the problem itself

[wind vault]

.close

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[short salmon]

(of course, it's always good to be aware that this assumes your functions are integrable and C^1 or smtg i think)

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[hardy venture]

hello, can someone help me with 65 and 66?

[random escarp]

,rcw

[random escarp]

,rcw

[random escarp]

,rcw

[hardy venture]

and at 66 I need to find the images of the functions

at 65 I need to find m in R, so functions are monotonous on I

idk what to do next

any ideas?

[calm coral]

@hardy venture Has your question been resolved?

[hardy venture]

<@&286206848099549185> - sorry for tag

[lucid oxide]

Ehhh, isn't it just for any m greater or equal 2, its monotone

[hardy venture]

but how do I write this mathematically

we aren’t allowed to use tools like this in class

😕

[lucid oxide]

Maybe say for function |x-m| there exists vertex on x=m s.t for all x less then 2 in I \in (-infty, 2) there exists vertex in I

[lucid oxide]

It's just a demonstration to show you

[hardy venture]

I guess it about the same for b), different numbers

what about ex. 66?

[lucid oxide]

For the range, ehh for a) it depends on y=|x| so we rewrite that into y, then f(0) = -1, f(y)_{y \to infty} = 1, while f’(y) > 0 S.t strictly increasing on [0, infty)

[-1, 1)

[hardy venture]

I’m in ninth grade, sorry, didn’t mention

don t know calculus

[exotic falcon]

That's amazing

[hardy venture]

I mean, I know a bit

but I m sure I can t use it

[exotic falcon]

What have you tried ma'am?

[hardy venture]

ma’am

😭

I’m a male

[exotic falcon]

oh my apologies sir!

What have you tried for this doubt

[hardy venture]

nothing so far

[exotic falcon]

Well

that's not a great start

[hardy venture]

oof

I missed the last 3-4 classes

that s the reason I have no idea how to solve them

I went to a hackathon

:)

[calm coral]

@hardy venture Has your question been resolved?

[calm coral]

@hardy venture Has your question been resolved?

[vague bison]

please use ,rccw for one counterclockwise rotation of 90° instead of three clockwise rotations of 90°

[vague bison]

but u can sketch the graph (in your mind, without drawing that out on paper)
that's sth u can do in tests/exams

[calm coral]

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[tacit basin]

I have a linear transformation T:$\mathbb{R}^3$ -> $\mathbb{R}^3$ and a matrix A with respects to a basis B. Does that mean that both the input and the output are coördinates of the basis B? Because up to now everytime there was only one basis in my book it said 'with respects to a basis B (considered twice)'

[potent lotus]

Jolqn

[tacit basin]

*a matrix A of the linear transformation T with respects to a basis B

[glass heart]

yes both the inputs and outputs are coordinates with respect to B

[tacit basin]

Thankyouu

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[civic pine]

Hello, I have a math exam today and I am having really bad trouble understanding which ones I need to worry about. He sent me an email. I need help on mostly everything. I DONT expect a lot of help I just need help understanding which math questions I need to do and work on. I’d also need help on my word problem questions, this is simple grade 10-11 math but I SUCK as I barely paid attention. I’ll be sending pictures of my word problem questions, I need help understanding how to do it more.

[civic pine]

Can pick any question, I just need help understanding what to do first.

[void umbra]

I reckon qn 28 is the simplest

[civic pine]

Would you help me with the simplest question and the hardest?

[void umbra]

So in the drawing the actual parts are 5 times longer than that in the image

As shown by the scale 1:5

So if we are given the length of an actual part we can calculate it's length in the drawing by dividing by __

[civic pine]

I’m lost

[void umbra]

So for example if a part of 5'' were drawn, the actual length of the part would be 5''*5=25''

In this case we're trying to find the drawn length of the part given it's actual length

So we want to reverse the multiplication by 5

How would we do that?

[civic pine]

25?

[void umbra]

Yes

[civic pine]

so the answer is 25?

[void umbra]

No

It was an example

[civic pine]

Oh

I never realized how much I never paid attention in math until now.

[void umbra]

Have you heard of division

[civic pine]

Yeah

Its fractions that mess me up and word problems

would it be 3? Like 15 divided by 5?

[void umbra]

Yea

[civic pine]

OH okay

Could you possibly help me with another one?

[void umbra]

Uh ok

[civic pine]

Okay, I’ll send another photo as it’s on another page.

question 40?

[void umbra]

Never heard of a grade before

Apparently it's a measurement of slope based on Google

[civic pine]

grade I’d say for us is like a slanted form

[void umbra]

So we want to convert this grade from a unit of in/ft

[civic pine]

idk how to explain really

[void umbra]

Into just a pure ratio of the lengths

So uh based on Google 1 foot is 12 inches

[civic pine]

yeah it’s 12 inches

[void umbra]

So uh how many inches are 32 feet

[civic pine]

384

[void umbra]

Yea

So 7/32 in/ft = 7/384

Then we multiply by 100 to get the percentage

So 7/384=700/384%

From here we do long division

Or if you're allowed to, a calculator

[civic pine]

calculator we got

Can use it

[void umbra]

Alright so go ahead and use that

Around to nearest hundredth

[civic pine]

What’s the “/“ for?

[void umbra]

Division

[civic pine]

Alright I kinda figured

I got 1.822916667

Haha I forgot to add “%” so now I have 182.2916667

[void umbra]

Yea nice so 1.82%

[civic pine]

So it would be 1.82%?

It is

[void umbra]

Yea

[civic pine]

okay

thank you, I’m getting it a bit now

Could we do another question?

[civic pine]

Wait, I kinda realized 7 is inches, 32 is the feet. 32 feet X 12 inches = 384% so we would need to do 7 / 384 % . So the 7 inches ontop stays there until we can divide it

[void umbra]

No 32 feet= (12*32)inches=384 inches

The 384 isn't a percentage either

[civic pine]

HAHA

I thought I was onto something

[void umbra]

Anyway you do have the answers right

Does the answer key come with workings or something

[civic pine]

What do you mean by that?

[void umbra]

Do you have the answers to these questions

Because this looks like an assessment book

Oh nvm

Yea it literally says there that there's an answer key mb😭

[civic pine]

😂😭yeah there is answers but I second guess myself so bad and I do need help understanding as trying to understand on my own IS ASS.

.close

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[steep juniper]

can someone try solving this problem that i made(uses derivatives specifically implicit differentiation):
7x^2 + 4y / x^2 - 5y = 0
Somehow i couldn't solve my own problem lmao

[leaden marsh]

like to find dy/dx?

[steep juniper]

yea

[leaden marsh]

which term are u having trouble with

the outer terms are power rule right

[glass heart]

I assume you wanted to write (7x^2+4y)/(x^2-5y) ?

$\frac{7x^2+4y}{x^2-5y}=0$ ?

[steep juniper]

like the part after deriving all of it which turned to
14x^3 - 35xdy/dx + 8xy - 20ydy/dx + 14x^3 + 4x^2dy/dx - 70xy - 20ydy/dx over (7x^2 + 4y) = 0

= 0

that and add = 0

[potent lotus]

Denascite

[steep juniper]

yes

[steep juniper]

it got me confused on this part

first i thought of multiplying both sides to remove the denominator and factor out the y primes and divide both sides by the numerator that is inside the parenthesis

is that correct or no?

[leaden marsh]

wait why is your denominator (7x^2 + 4y)

[steep juniper]

wait actually now that u mention it

omg i forgot

the denominator is (x^2 - 5y)^2

mb

[leaden marsh]

👝

[steep juniper]

anyways it should be:
14x^3 + 4x^2yprime - 70xy - 20yyprime - 14x^3 - 35x^2yprime + 8xy - 20yyprime and the denominator is (x^2 - 5y)^2

what should i do after it?

[leaden marsh]

mind your brackets

it should be
$\frac{14x^3 + 4x^2y’- 70xy - 20yy’ - 14x^3 + 35x^2y’- 8xy + 20yy’}{\left(x^2 - 5y\right)^2}$ technically

[potent lotus]

blahaquil

[leaden marsh]

then u factor

(specifically y’)

[steep juniper]

after that?

[leaden marsh]

then you just

move the non y’ terms to the other and divide as usual

[steep juniper]

so u move all since all y' were factored out?

[leaden marsh]

well some have the y’ factored out yea so u don’t move those terms

like

you’ll have (something)/(x^2-5y)^2 + y’(something)/(x^2-5y)^2

[steep juniper]

dang i dont get it

[leaden marsh]

like in normal implicit differentiation after you get the form with yprimes

what do u normally do

[steep juniper]

usually after getting the y primes u factor them out and divide the remaining values to the other side

[leaden marsh]

yea

[steep juniper]

how can u divide them out tho since there's still (x^2-5y)^2

as the denominator

[leaden marsh]

wait I’ve been going as if that’d be on one side only

yea multiply

by the denominator

[steep juniper]

what

so y'(something)/(x^2-5y)^2 = 0
u multiply both sides by (x^2-5y)^2 and after that u divide both sides by (something) and answer would be y'= (x^2-5y)^2 / (something)?

actually nvm that's wrong

hmm o wait nvm i get it

[steep juniper]

u multiply both sides by (x^2-5y)^2 ?

[calm coral]

@steep juniper Has your question been resolved?

[calm coral]

@steep juniper Has your question been resolved?

[calm coral]

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[nimble portal]

How can you get started with this?

[potent smelt]

This seems like an Olympiad question?

If so, then there is almost certainly a slick solution

Typically, for questions like this, the exact number of sides do not matter provided they satisfy some property, which in this case is probably 1 mod 4. So I would first try an analogous problem using a pentagon, and then prove that as long as you have 1 mod 4 the ratio of white vs black is the same. Finally, consider the case where we let the sides go to infinity, and we can see that in the case of a circle, it would have to be 1/2.

But I could be completely off base!

[nimble portal]

It’s from a Olympiad qualification comp, USAMTS

[nimble portal]

Will try this thanks

[leaden marsh]

keep in mind point B with the pentagon as well

[calm coral]

@nimble portal Has your question been resolved?

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@nimble portal Has your question been resolved?

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[final salmon]

Recursive → Explicit
how would i write the explicit formula for the sequence given recursively.

[fickle wigeon]

write the first few terms you will see a pattern

[final salmon]

its always plus 3

[fickle wigeon]

otherwise you can replace n+1 with n and n with n-1

so you get an = an-1 + 3 does that ring a bell

[final salmon]

but how can i do an = an + 2

or is that right?

an = 1 + (n-1)*3

right?

[calm coral]

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[somber swift]

can someone explain to me how you can use double integrals to calculate the average value for a function, theoretically.

[bronze adder]

I mean if you have a region $D$ with area $\mathrm{ar} , (D)$, then the average value of your function over $D$ would be
[ \frac{1}{\mathrm{ar} , (D)} \iint_D f(x, y) \dd{x} \dd{y} ]

[potent lotus]

neon

[somber swift]

So we get a value from the integrals and then divide by the total area?

[bronze adder]

yes

[somber swift]

Makes sense, thank you.

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[calm coral]

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[silver blade]

lets say we have a set of data, set A, and a set of coefficients, set B. Now consider the following:
(B1)(A1) + (B2)(A2) ... + (Bn)(An) = C
Now, elements of set A are both negative and positive. additionally, the elements of set A are extremely different, as in A1 could be -0.71 and A2 could be 732,000,000. also, the value if C is already known, meaning the goal is to find a set B that would make the equation true
in theory, elements of B should add up to 1, basically meaning each BnAn is the "percentage" of C thats made up from An, but because An can be pos or neg, not sure how this would work
Now, to add the last layer of complexity, this equation needs to hold true add up to C with a small error for multiple different scenarios while keeping set B the same. In other words, A(B)=~C1, and A*(B)=~C2
how would you determine set B to always roughly sum to C with the least uncertainty?

(this isnt an hw problem or anything, just an intermediate step im trying to figure out in a personal coding project. im currently taking linear algebra and also differential equations and it seems like it involves matrices so feel free to use matrices and if i dont understand a symbol or a transformation then ill just ask)

(also, if this should be put in one of the early uni/advanced math categories, since my end goal is coding whatever method to find set B, then lmk to move this)

[calm coral]

@silver blade Has your question been resolved?

[calm coral]

@silver blade Has your question been resolved?

[silver blade]

<@&286206848099549185>

[leaden thunder]

what does "set B to always roughly sum to C with the least uncertainty" even mean

"the goal is to find a set B that would make the equation true in theory, elements of B should add up to 1,"
how do you even know such a B exists without giving the range of A and value of C

for simplicity, just set n = 3

[silver blade]

if you have 2 sets of A's and 2 C's, its not possible for a set B to sum exactly to C in both scenarios, since B is the same in both scenarios. so i meant having the least total % difference between what actually is summed and what C is

[silver blade]

in theory, it would have to exist, no? lets say A1 is 500 and A2 is .5, and C is 1. then B1 = .01 and B2 = 1. the difficult part is finding those values when you introduce a 2nd set of A and value of C. it wouldnt be perfect, but that goes back to finding the least % difference between what AB sums to and what C actually is

[leaden thunder]

what do you mean "it would have to exist"

you said the elements "should add up to 1" but your example it doesn't

[silver blade]

Sorry yea that was wrong

In that example it should be B1 is .001 and B2 is 0.999, and they get as close as possible to 1

thats a simple system of equations. but my confusion starts when you add another series of A and another value of C

would you need to normalize A and C first?

[leaden thunder]

you'd have to even first define the problem coherently

you can do that by presenting a coherent n=3 example

and conditions on the vector A and the number C

[silver blade]

{A}1 = 723, -3.4, 1386
{A}2 = -682, 4.7, -986
C1 = 7.6
C2 = -4.3

(A1.1)B1 + (A1.2)B2 + (A1.3)B3 = C1 + error1
(A2.1)B1 + (A2.2)B2 + (A2.3)B3 = C2 + error2

What is {B} with |error1| + |error2| ≈ 0?

[calm coral]

@silver blade Has your question been resolved?

[calm coral]

@silver blade Has your question been resolved?

[calm coral]

@silver blade Has your question been resolved?

[calm coral]

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[clever plover]

can this question be done using modulo?

[clever plover]

if so then how

[mortal orbit]

What are the possible values of (...)² mod 3

[clever plover]

0,1

[mortal orbit]

Ok

[clever plover]

tried it that way, and a possibility arises when x2 y2 are both 0

and so is 3z2

[mortal orbit]

So show that if x or y is different from 0

[clever plover]

what does that imply? i dont understand

[mortal orbit]

Then there exists a solution (x',y',z') where at least one is not divisible by 3

[clever plover]

i rlly suck at proofs, mind showing me how? 😭

[mortal orbit]

But yeah, forgetting about the solution (0,0,0) is kinda bad

[clever plover]

I tried infinite descent on it, and thats the only way i could do it

but took a long tim

so I wanted to try this method but dont understand how

[mortal orbit]

It is linked to infinite descent though

[clever plover]

no way to do it w/o it?

[mortal orbit]

Suppose one of x,y,z is not 0

Let a,b,c be the 3-adic valuations of x,y,z

Then taking d = min(a,b,c)

And x' = x/3^d, y' = y/3^d,...

then x'² + y'² = 3z'²

If x' or y' is not divisible by 3

Then x'²+y'² is not 0 mod 3

Otherwise, both x',y' are divisible by 3, but z' isn't

So LHS is divisible by 9

But RHS isn't

end of proof

[zenith summit]

=> 3 | (x^2 + y^2) => 3 | x and 3 | y => (3x')^2 + (3y')^2 = 3z^2 => x'^2 + y'^2 = 3z^2/9 = 3z'^2 (where z' = z/3 < z assuming none are zero)

[clever plover]

k gimme a sec to go thru and understand it

.close

[calm coral]

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[clever plover]

got it, ty <3

[calm coral]

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[random escarp]

$\int f(x)$ dx

[potent lotus]

SELVATOR

[random escarp]

I tried $\int 2f'(x) - f"(x)$ dx

[potent lotus]

SELVATOR

[random escarp]

I got $\int -xe^x -3e^x$ dx

[potent lotus]

SELVATOR

[calm coral]

@random escarp Has your question been resolved?

[random escarp]

<@&286206848099549185>

[random escarp]

<@&286206848099549185>

[random escarp]

<@&286206848099549185>

Bruh

[calm coral]

@random escarp Has your question been resolved?

[random escarp]

We have f(x) = (1-x)e^x
And f''(x) - 2f'(x) + f(x) = 0

We need that integral

<@&286206848099549185>

[leaden thunder]

use integral properties then integration by parts

,tex .int sum

[potent lotus]

riemann

[random escarp]

Doesnt work

We dont have F(x)

[leaden thunder]

that's why i said integration by parts

[random escarp]

I think we need to use that expression from a

[leaden thunder]

have you learned that yet

[random escarp]

U mean \int f(x) + g(x) = \int f(x) + \int g(x) ?

[leaden thunder]

this is just sum rule i stated above

If $F'(x) = f(x)$, then $\int f(x) dx = F(x) + C$ by fundamental theorem of calculus

[potent lotus]

riemann

[random escarp]

Yes

How to use it here in our case

[leaden thunder]

use $a(b-c) = ab - ac$

[potent lotus]

riemann

[leaden thunder]

on this

[leaden thunder]

this is called distributive property

[random escarp]

Yes, e^x is integrable

But xe^x is not

[leaden thunder]

.

[random escarp]

xe^x - \int e^x ?

[leaden thunder]

,w int x e^x

[potent lotus]

Results provided by WolframAlpha

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

Requested by rie.mann

[leaden thunder]

looks right

[random escarp]

Idk so

Ty

.close

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[jolly plover]

hello how do i derive cube roots above?