help-42

[glad sinew]

but the only reason I set delta < 1 here is because its a fraction

[upper sparrow]

Not quite no, you can restrict delta to be less than whatever [positive] number you want whenever you want

The fact that you had a fraction last time is irrelevant (also that argument I find a bit sus too but another time )

[glad sinew]

ah ok, cause you are just setting the distance for delta... like as long as you can find the sorresponding eps and eps+f(a), eps-f(a) you are good

so as long as delta is positive

and bigger than 0

i think you can set delta to be 0 too, if its just a point, but not sure how that helps with continuity

[glad sinew]

nah now I gotta know

ok yea

yayy

working on this one now, Im thinking to prove by contradiction, negate the continuity def, use triangle inequality (negative one) when dealing w inequalities and hopefully find smth that contradicts?

[upper sparrow]

Yep, you can set your delta to be whatever you want and that makes your life easiest

[upper sparrow]

And cool happy with that, there's an implicit intermediate step where you use the triangle inequality I see

[upper sparrow]

And for this, it should be much easier and can be done directly...

[glad sinew]

yea I think I can just say that ||f(x)| - |f(a)| | <=.... and then directly just say its continuous

like ofc I had all the rest of the continuity definition

I crossed out the part where I was trying to do contradiction

hm maybe I should rewrite?

ok i’ll be back in 30min-1hr

[upper sparrow]

Yep, reverse triangle inequality is all you need

[upper sparrow]

And you should

[calm coral]

@glad sinew Has your question been resolved?

[glad sinew]

okiee

i just realized for converse if |x-a| |is equal to | |x| -|a|| then the delta doesn’t hold

but there still exists a delta where it could be the case that:
| |x|-|a| | < delta < |x-a|

for the converse

[glad sinew]

for this question I dont get how max and min of a function affects the function's continuity if you already know that f and g are continous

<@&286206848099549185>

[glad sinew]

why is f increasing a reason for it to be bounded by f(a)?

<@&286206848099549185>

[glad sinew]

i dont get it :((

I get the part about a-1 <= x < a

right

but you are saying f is continous at a, so theres also x bigger than a

right? as long as the x is within (delta - a, delta+a)

<@&286206848099549185>

[random scaffold]

Can you send the entire proof

[glad sinew]

hey i’m actually gonna sleep now

i took screenshots of my question

i’ll ask again tmr

thank you tho

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[radiant notch]

can someone help me with this:
solve for the Fourier Series in Trigonometric and Exponential Form

[calm coral]

@radiant notch Has your question been resolved?

[radiant notch]

<@&286206848099549185>

[pallid halo]

what have you tried?

[calm coral]

@radiant notch Has your question been resolved?

[radiant notch]

nothing i dont understand about it lmao

[pallid halo]

probably you have some formulas you could start with?

[pallid halo]

ok

time to start integrating

[calm coral]

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[calm coral]

Please don't occupy multiple help channels.

[mossy cliff]

help with all quétions

<@&286206848099549185>

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[ebon narwhal]

can someone help me with b and d

[warm warren]

Do you have formulas

[ebon narwhal]

i know the base formula

for compound interest

[warm warren]

What have you tried then

[ebon narwhal]

for b i tried

5000(1+{0.01/12})^72

[lusty quest]

You’re applying 1% compound interest 12 times every month for 6 months?

[ebon narwhal]

i dont believe so

im applying 1% compiund interest onto 5000 every month for 6 years

[lusty quest]

The 12 means apply interest 12 times

Per month

[ebon narwhal]

oh

im just trying to find the balance after 6 years of 1% per month

but not sure how to do that

[lusty quest]

Ah

So ignore the n value

Cause that’ll always be 1

Now find how many months in 6 years

[ebon narwhal]

72

[lusty quest]

And that’ll be your t

5000(1+{0.01})^72

That should be the right answer

[ebon narwhal]

ah i see

[lusty quest]

Another thing

The equation is

P(1+r/n)^nt

So your original one was missing the final n in the exponent too

[ebon narwhal]

whats n standing for?

[lusty quest]

(Should’ve been 72*12)

n is the amount of times interest applied per time period

[ebon narwhal]

i see

[lusty quest]

So I could apply the interest 3 times a year or something

Yeah

[ebon narwhal]

so id onyl do this if the question were asking how much interest is he gaining per month or something

[lusty quest]

No

[ebon narwhal]

without the ^72?

[lusty quest]

For that you would find how much he has after a month and subtract how much he had the previous month

[ebon narwhal]

oh i see

[lusty quest]

Phhh wait

Did you divide by 12 because you wanted to find the percent per month?

[ebon narwhal]

yes

[lusty quest]

Ah

I see

[ebon narwhal]

but i jsut read the question again and it asks for the balance

[lusty quest]

Yeah one period is already a month, so that’s not needed

[ebon narwhal]

so this is right

[lusty quest]

Yes

[ebon narwhal]

makes snsese

thank you

[lusty quest]

No problem

[ebon narwhal]

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[blazing coyote]

$\int_0^3\left[x^2\right]+\left[\frac{x^2}{2}\right]dx$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

other than breaking this into multiple intervals

is there an way too solve it

for instance

$\int_0^n\left[x\right]dx=\frac{n\left(n-1\right)}{2}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

is there any general closed form for this too

where$[x]$ is $\floor{x}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

I don't think there is a similar closed form

For instance consider [x^2] upto 1, it's 0

Similarly the integral to 2 is

0+sqrt2-1)+ something

Which suggests a neat closed form doesn't exist

Am I right?

,w integral from 0 to n of floor (x^2)

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[blazing coyote]

<@&286206848099549185>

[dense rapids]

break the integral

at integral

values

and then add

[blazing coyote]

yeah, I know that

but is there a general closed form

[dense rapids]

for x^2

i dont think so

[blazing coyote]

Thanks!

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[honest swallow]

its simple but i cant seem to understand

[hasty fiber]

drawing a venn diagram might help

[honest swallow]

it isnt really specified if theyre disjoint or not so i am confused

[hasty fiber]

you'll get the same result regardless

you can draw two diagrams if you want to consider both cases

[honest swallow]

alright ill try

but are you sure any property or law wont apply here?

like maybe we can use DeMorgans law somehow?

[hasty fiber]

if you can rewrite X-Y in a form that allows you to invoke de morgan, you can certainly use it

[honest swallow]

hmm alright

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✅

[urban gull]

Someone prove that ô1 is 2*angle m1

[calm coral]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[honest swallow]

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[potent lotus]

Result:

5

[serene harbor]

like like

i dont get how euler totient function works here

i just followed blindly

nd got remainder to be 5

also but i made an assumptio

that after 6 sets remainder remain same

wait

https://math.stackexchange.com/questions/4592762/find-the-remainder-of-333-3-divided-by-46

i did not understand this part

[knotty blade]

Okay

So there’s a theorem in number theory that says a^phi(n) is congruent to 1 mod n

When a and n are relatively prime

[serene harbor]

euler totient

[knotty blade]

This is Euler’s theorem, and it’s not awful to prove at all in a reduced residue system

[knotty blade]

Yes

Do you know Fermat’s little theorem?

[serene harbor]

i did get that
but i dont understand this last part

[serene harbor]

nope

but ik we predict prime numbers with it? ig

[knotty blade]

Okay so

Are yoy familiar with modular arithmetic

We may have to do some background building

[serene harbor]

a lil
i did not touchhed upon NT much

we usually do remainder ques with binomial thm

[knotty blade]

Okay

Well basically modular arithmetic is how you get here

So did you understand all lol the previous steps

Or no

[serene harbor]

i kinda swallowed the concept of euler totient function
i just dont know why we using it

usually i used the function to calculate number of positive integers less than or equal to n nd coprime to it

[knotty blade]

So a little bit of background building g

We say b is a inverse of a mod n if ab = 1 mod n

[serene harbor]

example

[knotty blade]

2 and 3 mod 5

2*3 = 1 mod 5

[serene harbor]

ohh the negative remainder thingy

[knotty blade]

Ok

So I’m going to make a claim here that I’m not going g to justify bc it would take forever

If a is relatively prime to n, then a has a multiplicative inverse mod n

Reasonable?

[serene harbor]

so thats the reasoning to using ET funcn

[knotty blade]

In a sense

Does that statement sound ok

[serene harbor]

oh a lil
i'll do a research myself after

i get after
but the last line

why more than 4
nd it stays 41 mod 46

will in case of 5
it will stay same after tower of 6

[knotty blade]

Okay s

[knotty blade]

Well can those 4 and 5 cases really exist?

Okay

So the thing is

We KNOW that x3 is 3 mod 4

Therefore, we can compute what x2 is going to be

And what x is going to be

Right?

[serene harbor]

here

[knotty blade]

It’s always going to be 3 mod r

Because 3^k is always 3 mod 4

For odd k

[serene harbor]

huh

for odd

oh

[knotty blade]

This is only true when we go 4 or more steps up though

Because we need to get to x3 first

So we need x0, x1, x2, x3 to all exist

[serene harbor]

so in my question
can we say 6 steps
since it comes like that at 6

[knotty blade]

Yeah

X3 would be 3 mod 4

And then you keep going down

[serene harbor]

i followed that
i got 5 to be the remainder

[knotty blade]

How

You did an error somewhere

[serene harbor]

this question

i got 5

[knotty blade]

Ok so your issue here is that 5 and 24*5 aren’t relatively prime

So Euler’s theorem doesn’t work

lol

[serene harbor]

so how do i go bout this

😭

[knotty blade]

Well what you need to do is divide out 5 from the modular congruence

So you get 5^( (5^5^5^5…)-1))

Let x0 be this mod 24

Let x1 = 5^5^5^5…-1 mod phi(24)

And keep going

[serene harbor]

ohhh

lemme note it out nd try

i found the period to be 8

[knotty blade]

What?

💀

[serene harbor]

[knotty blade]

Sully

Bye bye

I have things to do now

[serene harbor]

rip

oh ok t

thanks tho

.close

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[sturdy gyro]

Can someone check whether its correct or not? 😭

[near crag]

You can plot the function you get on desmos.com/graphing

[sturdy gyro]

This question need to plot? 💀

I thought u just find the value

[minor knot]

uhh

[near crag]

Plotting is a good way to check your answer

[sturdy gyro]

Ohh

[minor knot]

these are wrong

[sturdy gyro]

Ohh kk

I understand the 4 but why 3

[minor knot]

(x+3)² is a shift to the left

and the +4 is a shift upwards

[sturdy gyro]

Ohh kk

[minor knot]

yea thats whats wrong

[sturdy gyro]

Tqq

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[slate grail]

We have a,b,c are real numbers, a,b,c>=0 and a+b+c=3
Find the minimum value of P = 1/(a+1) + 1/(ab+1) +1/(abc+3)
Sorry I dont know how to use latex

[remote mural]

I am a bit confused about how you are meant to compare Taylor approximations together. Consider a function $\6fx$, which is analytic on its domain, with two taylor series representations, $\6{T_c}x$ and $\6{T_k}x$ each centered around some points $c$ and $k$. Now, with the functions being analytic, it follows that $\6fx = \6{T_c}x = \6{T_k}x$. But if you are looking to make some approximation of the function $f$ via its Taylor series of order $n$, then it is not generally true that [
\6{T_c^2}x = \6{T_k^2}x
]
So my question becomes how exactly do taylor approximations with different centres compare with each other? What exactly is the error that makes us not able to have the series be equal to each other?

[remote mural]

v

uh

[slate grail]

We have a,b,c are real numbers, a,b,c>=0 and a+b+c=3
Find the minimum value of P = 1/(a+1) + 1/(ab+1) +1/(abc+3)
Sorry I dont know how to use latex

[dusky crescent]

can anyone guide me through the steps to match the equation of a slope field. This one is either xy, x, x - y, or x + y

the one that I am confident to rule out myself is the x

I also notice that the slopes are positive all in first quadrant and negative all in third quadrant

and that slope is 0 in the middle there

[remote mural]

help channels r broken

open a forum thread in #1021175428326633542

[dusky crescent]

Ohh okay thank you!!

[indigo gust]

What's wrong in my solution?

[spice prairie]

I have a question to find the sum of the first 9 terms of the series 6+0.6^26+0.6^3+60.6^46+... Can u please check if im correct: I got that r = (0.6^36)/(0.6^2*6)=3/5 the S9=(6(1-3/5^9))/1-3/5 = 14.85

[humble zealot]

Hi guys, can somebody please help me understand what are F^S and R^[0,1]? I get that for example R^2 is the set of all the ordered pairs of real numbers and if I understood correctly it denotes the set of functions that sends each pair (a,b) where a,b are in R to a number in R. Basically, every point on the plane is sent to a number on the real line.

[silk scroll]

i have a question

Jack is reading a book. If on the first day he reads 3|7 of the pages of the book and on the second day he reads 3|10 of the remaining ones, calculate how many pages there are in the book knowing that there are still 80 pages left to read. can someone help me??

[unkempt drift]

So what fraction of the book is left?

[silk scroll]

4/7

[unkempt drift]

Well, 1 - 3/7 = 4/7

But you still haven't considered the 3/10

[silk scroll]

so?

[unkempt drift]

(btw he reads 3/10 of the remaining pages)

[unkempt drift]

So 4/7 isn't correct

[silk scroll]

3/10 of 4/7

[unkempt drift]

Yeah, you're nearly there

So what fraction of the remaining pages does he not read?

[silk scroll]

19/70

[unkempt drift]

Nope, it's not 1 - 3/7 - 3/10

3/10 is a fraction of the remaining pages

[silk scroll]

4/7-3/10 ??

[unkempt drift]

Stop guessing

[unkempt drift]

Okay, so John can either read the remaining pages, or not read the remaining pages

So what do these two fractions have to add up to?

Given that they are the only two options

[silk scroll]

3/10 and 3/7

[unkempt drift]

Ignore the 3/7 here for now

Cause we're talking about a fraction of 3/7

[silk scroll]

we have to add 3/10

[unkempt drift]

Okay, so these two parts have to add up to 1 - 3/7 = 4/7

[silk scroll]

yeah

thank u

[unkempt drift]

So 3/10 * 4/7 + 7/10 * 4/7 = 4/7

Or just 1 - 3/10 is 7/10

That's what I was looking for

[silk scroll]

7/10

[unkempt drift]

So he hasn't read 7/10 of the remaining pages, or 7/10 * 4/7 in total

Yeah so it's just 7/10 * 4/7 * x = 80

[silk scroll]

28/70 *x =80

[unkempt drift]

Well the 7s cancel

$\frac{7}{10} \frac{4}{7} x = 80$

[potent lotus]

south

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[mental cosmos]

How do i do this?

[torpid canopy]

See which one violates properties of an equivalence relation

[mental cosmos]

yes i tried that but im not sure how to do it without being given two sets

[potent igloo]

!show

[calm coral]

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

[potent igloo]

what do you mean "given two sets"?

[mental cosmos]

I found its reflexive given k=1

[potent igloo]

What is "its"?

[mental cosmos]

R

[potent igloo]

sounds good so far.

[mental cosmos]

For S when k=0 its reflexive

[potent igloo]

yup

okay, so they're both reflexive.

Are they both symmetric?

[mental cosmos]

but i dont know how to check if they are symmetric

or transistive

[potent igloo]

Assume xRy is true, does this imply that yRx is true too?

[mental cosmos]

No

Actually yes

[potent igloo]

Can you provide an example?

[mental cosmos]

For 1/k

[potent igloo]

Or, prove why yes?

[potent igloo]

But is 1/k always an integer?

[mental cosmos]

Oh no it is not

So it is not true

because if x=1 and y=2 it works

but for y=2 and x=1 it does not

[potent igloo]

correct

and what about the relation S?

[mental cosmos]

And the relation S is reflexive

Because for x=1 and y=2 it works for k=1

and for x=2 and y=1 it works for k=-1

[potent igloo]

exactly right

[mental cosmos]

Amazing, thank you so much 🙂

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[orchid slate]

how do u do this question

[pure kayak]

start from f'', make a function that satisfies the conditions for that

then you can integrate and solve for the two unknown constants using the rest of the info

[orchid slate]

this is the f"

graph

\

<@&286206848099549185>

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[lunar token]

help with understanding this proof

[calm coral]

@lunar token Has your question been resolved?

[calm coral]

@lunar token Has your question been resolved?

[untold compass]

the idea is that every domino on a checkerboard takes up a black and white square, so when you have a board with more white squares than black squares, it can't be done

because anything with dominos will cover an equal number of black and white squares

[calm coral]

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[void pendant]

Given a 5 6 Matrix, in how many ways can we move from cell 1 1 to 5 6, if we are allowed to move right or down and we can only change directions three times

[void pendant]

I tried 9!/(4!5!) which is 126 but I'm not sure how to check for the cases where it changes directions more than 3 times

The correct answer is 33

[civic dirge]

You can stars and bars the moves

[atomic palm]

id break it into two cases, since the direction change amount is odd, if we start with an UP, our ending move must be a RIGHT

[civic dirge]

A star being "move forward"
And a bar being "turn"

Ah, but that doesn't count for less than 3 turns

Or it doesn't count for leaving the matrix

[atomic palm]

it is definitely some form of stars and bars

[calm coral]

@void pendant Has your question been resolved?

[atomic palm]

@void pendant what do you need help with?

[void pendant]

I was wondering if I could adjust the formula

So i dont take cases

[atomic palm]

u probably dont need cases, its a multiplication of two binomial coefficients

stars and bars on two things, the UP's being distributed into 2 blocks, and RIGHT's being distributed into 2 blocks

[void pendant]

How many stars and bars?

5 stars and 4 bars?

Thats for 3 changes of direction,right?

[atomic palm]

yeah

[void pendant]

I did xD yJ (5-x)D (4-y)J where x<5 and y<4

Sorry for notation but d is right and j is down

For x there are 4 cases multiplied by 3. And then I do the same when starting with a down

nvm I think I got it

thanks

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[unique fiber]

Hi, the Picture on the left is the question, the right is my work so far. I’m not really sure what to do next, and I also don’t know how these planes are represented visually.

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@unique fiber Has your question been resolved?

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[glad sinew]

HIIIIII

[glad sinew]

Im back

lock in time

[glacial nacelle]

um

do you have a question?

[glad sinew]

hi well, for now no

[glacial nacelle]

then why do you have a channel?

there are discussion channels

[glad sinew]

cause its annoying to open, close, reopen channels for each question

[glacial nacelle]

so?

[glad sinew]

I usually have alot of questions as I work through the problems

[glacial nacelle]

then open the channel as you go

[glad sinew]

theres open channels tho

[glacial nacelle]

it is not that deep

[brazen elbow]

please only open help channel if you have one question

[glad sinew]

hmmm okok

.close

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[nimble urchin]

express as a product: 1-cot(a)

[nimble urchin]

@amber roost

[amber roost]

Basically you know

$cotx = \frac{cosx}{sinx}$
Put every term on sinx

[potent lotus]

Tittom_123

[amber roost]

So 1 become sinx/sinx

[nimble urchin]

alr

1-cot(a)
Csc(a)(sin(a)-cos(a))
cos(a)(sec(a)-csc(a))

these are the 2 forms i got

but i cant express them as a product

[amber roost]

the csc one is the right one, because it's a product

[nimble urchin]

lol but u cant use subtraction

[amber roost]

They are both products

[nimble urchin]

for 1+tan(a)

1+tan(a)
sec(a)(cos(a)+sin(a))
Sin(a)+cos(a)
Sqrt(2)(1/sqrt(2) * sin(a) + 1/sqrt(2) * cos(a))
Sqrt(2)(cos pi/4 * sin(a) + sin pi/4*cos(a))
Sqrt(2)(sin(a)cos(pi/4)+cos(a)sin(pi/4))
Sqrt(2)*sin(a+pi/4)
sec(a)*sqrt(2)*sin(a+pi/4)

[amber roost]

You can't simplify without the -

[nimble urchin]

the bottom equation is how u want it

[amber roost]

Where did the secx went?

[nimble urchin]

sec(a)

[amber roost]

Yes

[nimble urchin]

sec(a)sqrt(2)sin(a+pi/4)

this was the final equation

[amber roost]

And why √(2) ?

[nimble urchin]

i took it out of sin(a)+cos(a)

[amber roost]

How?

[nimble urchin]

Sqrt(2)(1/sqrt(2) * sin(a) + 1/sqrt(2) * cos(a))

it likes cancels out

[amber roost]

No, I mean from where does it appears

[nimble urchin]

i just factored it out

[amber roost]

Oh

[nimble urchin]

yeah its a weird question

[amber roost]

This doesn't work because

$\frac{1}{\sqrt{2}} \neq cos(\frac{\pi}{4})$

[potent lotus]

Tittom_123

[nimble urchin]

yes it does

1/sqrt(2) = sqrt(2)/2

cos(pi/4) = sqrt(2)/2

45-45-90 triangle rules

@amber roost

[amber roost]

You are better of working with 1/2

[nimble urchin]

ik but i got 1+tan(a) correct

just idk how to do 1-cot(a)

[amber roost]

Like this@nimble urchin

Since

$cosxcosy = \frac{1}{2}\left(sin\left(x-y\right) + cos\left(x+y\right)\right)$

$sinxsiny = \frac{1}{2}\left(sin\left(x-y\right) - cos\left(x+y\right)\right)$

[potent lotus]

Tittom_123

[calm coral]

@nimble urchin Has your question been resolved?

[nimble urchin]

@amber roost

its csc(a) for final equation right

[amber roost]

Oh, yeah, mbd lol

[calm coral]

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[nimble urchin]

thanks @amber roost

it was correct

[amber roost]

Didn't you had an other problem or nah?

[nimble urchin]

no one got this so far

[calm coral]

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[nimble urchin]

@amber roost

[amber roost]

$\sqrt{2} = sec(\frac{\pi}{4})$

[potent lotus]

Tittom_123

[amber roost]

Or sec45°

[amber roost]

$\frac{cos45}{cos10}\left(tan30 + cos180\right)$

I will let you work from this now

[potent lotus]

Tittom_123

[quaint sphinx]

? what

[amber roost]

What?

[quaint sphinx]

what is that expression

[amber roost]

This with trig

[quaint sphinx]

that is just extremely wrong

[amber roost]

Let me help my student, all the help I gave him yet gave him the right answers@quaint sphinx

[amber roost]

It isn't

[quaint sphinx]

first it helps to rewrite 3/sqrt6 as sqrt3 / sqrt2

[amber roost]

\rotate

[nimble urchin]

How’d u get this

[quaint sphinx]

from there you would find it useful to perform certain manipulations (i.e. factorization) as well as rewriting other expressions in terms of trig identities

[nimble urchin]

Alr thx

[amber roost]

Still, we arrive at the same result

[quaint sphinx]

dude you randomly replaced a sin10 with a cos10

don’t act all arrogant when helping someone when your work is just nonsense

[nimble urchin]

I’ll try to do it tomorrow

[amber roost]

Oh, yeah. I have done a typo, my bad

[nimble urchin]

@amber roost then it wont work

cause then u can't factor

[amber roost]

Without typo lol

[nimble urchin]

how'd u get that for the left expression

@amber roost

[amber roost]

Csc45 = sec45

@nimble urchin

[nimble urchin]

3/sqrt(6)*cos(10) becomes 3/sqrt(3)*sec(45)*cos(10)

then what

and where did the 3 go

[amber roost]

3/√(6) = (√(3)/√(2))

[calm coral]

@nimble urchin Has your question been resolved?

[calm coral]

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[minor island]

how are P1 and P2 negitive?

[pale blade]

Cause they're downwards what?

[minor island]

how do they know its a clockwise moment?

[lusty quest]

Find the pivot point

[pale blade]

Cause we're analyzing joint A

[lusty quest]

Then just imagine what direction it’ll be turning in if you pushed at that point

[minor island]

ah gotcha

[pale blade]

Consider pulling onto point B in the direction of P1, with A as your pivot

[minor island]

so when im doing my own work, when should I make clockwise or counter clock wise positive

[lusty quest]

You don’t need to

Just put all clockwise torques on one side

And counter on the other

[minor island]

am I over thinking this

[lusty quest]

Maybe lol

[minor island]

ok cool

tysm ill be back later lol

[lusty quest]

Counter torques = clockwise torques

Both are positive

[minor island]

gotcha

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[finite quest]

is b) correct?

i did it with my friend but i feel like the right answer would be

10sin(5x)cos(5x)

[strange lichen]

yea thats right

ur answer

but if u apply the sin(2x) identity

given sin(2x) = 2sin(x)cos(x)

we see that 10sin(5x)cos(5x) = 5 * 2sin(5x)cos(5x) = 5*sin(10x)

which is also the answer given in the screenshot

i dont think so

theyre differentiating btw

[finite quest]

soooo

[finite quest]

is both answers correct?

[glacial nacelle]

oops

mb

[strange lichen]

yes, theyre both correct

theyre equivalent

[finite quest]

okkk great thank you stephen🫡🫡

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[cloud wren]

hi there could i get my solution reviewed for q2

[cloud wren]

so what I did was first start off by finding r'(t) so I can solve int_c F(r(t)) dot r'(t) dt, then just went on to solve integral from there

[civic dirge]

The logic looks good. Did your answer get rejected or something?

[cloud wren]

nope i don't have the solution, just wanted to check if my steps were correct.

thanks

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[thin gulch]

what's your question?

[trim raft]

Solve the inequality

[trim raft]

@thin gulch

[thin gulch]

ok, what's your thought?

[trim raft]

For the function to be defined I have found the domain of the roots and it's -1<x>1

Now since both are positive, i have done a cross multiplication

(the denominators)

[thin gulch]

for sqrt(x^2-1) only, but yes that's the right start

[trim raft]

Yeah, the other one is always positive

[thin gulch]

ok so you have
$x \sqrt{x^2-1} > x \sqrt{x^2+1}$

[potent lotus]

lgkoo

[thin gulch]

also I think you meant to say x>1, and x<-1, not -1<x

[thin gulch]

without any explanation I actually cannot follow your work, not a single word is written, such as when/if to test different scenario. It's important to write down your thought because examiners and ppl cannot read mind and you don't want them to be confused.

[trim raft]

Okay i will try to put the words

[thin gulch]

ok I think I know what you're doing. So the hint here is: since we know sqrt(x^-1) is not defined for -1<x<1. Consider the two cases: x>1 and x<-1 separately

[trim raft]

Okay a minute

[thin gulch]

right idea but slight mistake

[trim raft]

What is it ?

[thin gulch]

you are right that $\sqrt{x^2-1} - \sqrt{x^2+1} < 0$ for x > 1

[potent lotus]

lgkoo

[thin gulch]

but then the product

$x(\sqrt{x^2-1} - \sqrt{x^2+1})$

[potent lotus]

lgkoo

[thin gulch]

is a product of positive and negative

[trim raft]

OMG yes, should have the same sign

x<0 not x>0

[trim raft]

@thin gulch

[thin gulch]

correct

[trim raft]

So the final answer will be the intersection of the domain of root and the answer we got now ?

[thin gulch]

yeah, which is just x<0

[trim raft]

But the book mentions the answer as -1>x>1

[trim raft]

Guess it's vogue ?

[thin gulch]

-1>x>1? That cannot even be true, how can -1>1?

[trim raft]

Sorry, i mistyped. -1<x<1

[trim raft]

@thin gulch

[thin gulch]

ok, but that's just the region where x/sqrt(x^2-1) is undefined. Can I see the question?

!original

[calm coral]

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

[trim raft]

Q. 11 @thin gulch

[thin gulch]

then yeah it's x<-1

[trim raft]

But how is x>1 ?

[thin gulch]

?

the inequality doesn't hold when x>1

[trim raft]

So, the book answer is vogue because the solution and graph both disproves it ?

[thin gulch]

uhh I'm not sure what vogue means, but the book answer is wrong

[trim raft]

Thanks, can you please mention the tool that you used to acquire the graph ?

[thin gulch]

ofc

it's desmos

https://www.desmos.com/calculator

[trim raft]

Thanks, problem solved

[thin gulch]

you're welcome

[calm coral]

@trim raft Has your question been resolved?

[calm coral]

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[half raven]

#29

[half raven]

i dont know where to start

i dont see an identity that i can use for the sqrt

[north grail]

that seems incredibly hard to integrate

cos^4(x)

[thin gulch]

try using the substitution of u = sinx

[pearl dune]

Optimistically I want to say that's a typo and it should be sqrt(1-sin^2) as then it's a super easy simplify and subtitute but maybe it's just obnoxiously hard lmao

[north grail]

yes i agree with taybee

[half raven]

uh what does this do

i also forgot to change limits of integration but thats ok

,rotate

[north grail]

u r cooked bro

that integral is demonic

[half raven]

😭😭😭😭😭

[thin gulch]

you can then write cos(x) in terms of sin(x) then in terms of u

[half raven]

idk what you mean

(1 - sin^2 x) cosx

[thin gulch]

$\cos(x) = \sqrt{1-sin^2{x}$

[potent lotus]

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

[thin gulch]

so $cos^3{x} = (1-\sin^2{x})^{\frac{3}{2}}$

[half raven]

how would i get it in terms of u

[potent lotus]

lgkoo

[thin gulch]

remember our substitution is u=sinx

[half raven]

ohhhh

youre so right

[thin gulch]

the result looks ugly but the process isn't too difficult once you see the right substitution to make

[half raven]

alright what now

[north grail]

how would u integrate ((1-u^2)^3/2)/√(1-u)

[thin gulch]

c'mon I've seen you helped ppl a lot, you can spot how to simplify this

[north grail]

hmm lemme see

[thin gulch]

hint: what does (1-u^2) factorise to

[half raven]

uh

[north grail]

LMAO its simple cancellation

[thin gulch]

😉

[north grail]

u r a genius

[half raven]

i forglorb

[thin gulch]

thx, but I'm really not XD

[thin gulch]

(1+u)(1-u)?

[half raven]

ohh

and then

uhh

cancel the bottom one

to getttt

like this?

,rotate

[thin gulch]

not quite

remember if I have x^(3/2) divided by x^(1/2), what does that give me?

[half raven]

x

so just (1-u)

instead of cubed

[thin gulch]

yup

so you have to integrate (1+u)^(3/2) * (1-u)

any idea what the method for this is?

[half raven]

uh integration by parts?

[thin gulch]

yup

[half raven]

okay

would you recommend u = 1-u

[thin gulch]

hmm I don't think it makes too much of a difference when doing the by parts

I feel like you're more likely to make mistake by making another substitution

but if you're confident, go for it

[half raven]

i’ll try

[pearl dune]

I think he means u as in one can write the parts formula as uv - int(vdu) rather than making a second u-sub

To which I would say, yes, that is the one I would differentiate XD

[thin gulch]

ohh

yeah absolutely, differentiate 1-u

[half raven]

i came to a conclusion

im not@gonna bother simplifying it all the way

so i just finished integration by parts

substituted the values

and boxed the answer

profs fault for giving such hard question 🗣️🗣️

[thin gulch]

hahaha honestly fair enough

[calm coral]

@half raven Has your question been resolved?

[calm coral]

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[hallow mountain]

am i missing something?

[hallow mountain]

cant i just plug in 7 and be done

[thin gulch]

yes

[hallow mountain]

does the + change anything

the one above the 7

[thin gulch]

not in this case

[exotic blaze]

That just means the limit approaches from the positives, it doesn't change anything in terms of this specific example

[hallow mountain]

ok thx

/close

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[hallow mountain]

kinda confused how to handle this

[lusty turret]

wot have u tried?

[hallow mountain]

i tried combining the fractions in the brackets

but in the end it got me to 3/x * 1

[lusty turret]

ok

that's not right

[thin gulch]

^

[lusty turret]

try combining the fractinos in the brackets again

[hallow mountain]

thats what i did

[lusty turret]

top is wrong

x-1 + x+1 is 2x

not x^2+1

[hallow mountain]

ok

i fixed it and got it to be -6

[calm coral]

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[keen grove]

what would be the proof for this?

[lusty turret]

I dunno if this is right

prolly smth easier

but u could let A be [[e,f],[g,h]] and then let X be [[a,b],[c,d]]

where a, b, c, d, e, f, g, h are in the reals

and then do matrix multiplication

[keen grove]

like just set arbitrary matrices?

[lusty turret]

ye

its disgusting but that's one way

[keen grove]

ok i will give it a go

[stoic oyster]

the nice way is to test for some specific matrices X only

if you plug in a matrix X, it will give you some equations (linear equations this time, so much less disgusting than just putting two arbitrary matrices like poggers) that the entries of A must satisfy

if you choose your Xs smartly, you can get to the result fast

[keen grove]

is what ive done here okay?

im not too sure where to go

[stoic oyster]

yeah that's the problem with going this way

the system is a pain in the ass

[keen grove]

wdym by test for specific matrices X only?

[stoic oyster]

our premise is that we have a matrix A such that for all X, AX = XA

we're free to pick whatever X we want, for example X = [1 0 ; 1 1]

and we know that A [1 0 ; 1 1] = [1 0 ; 1 1] A from our premise

this tells us some information on the entries of A

[keen grove]

so u can choose X to be whatever u want?

I thought the point was to show that it works for all real 2x2 matrices X

[stoic oyster]

you know that it works for all real matrices already

you don't have to show that

[keen grove]

ohhhhh

ahhh ok i see

ahh right that makes sense now

ty

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[limpid mulch]

How can I use both similar triangles as well as pythagoras theorem to get b)

[calm coral]

@limpid mulch Has your question been resolved?

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[stuck quartz]

someone break this down conceptually how to solve this using chain rule

[stuck quartz]

im used to doing things like (3x+1)^7 or (x^2+5x-6)^9

so tis quite confusing

[minor knot]

why did you open another channel

[stuck quartz]

i thought u were afk

[minor knot]

oh.

[stuck quartz]

Help renz plz

[minor knot]

going back

,,2e^{(1-x)^2}\cdot(-2)(1-x)

[potent lotus]

Renz

[minor knot]

you cant combine exponents like that

in that example, you can only combine 2 and -2

[stuck quartz]

OH

i forgot yeah

[minor knot]

,,-4e^{(1-x)^2}(1-x)

[potent lotus]

Renz

[minor knot]

thats the derivative

it cant be further simplified i believe

just multiply the 2 and -2 together, and put it in front

[minor knot]

now lets do these

[stuck quartz]

No, no

thats fine

i know how to do those

[minor knot]

ah oki

[stuck quartz]

im asking what was ur thought process for knowing what to do

[stuck quartz]

this specifically

im still kinda confused

chain rule is taking derivative outside, leaving inside function * taking derivative of inside function

[minor knot]

Ah alright leme give you an example

[minor knot]

with exponentials

[stuck quartz]

Ok

[minor knot]

$e^{x^2+2x}$

[potent lotus]

Renz

[minor knot]

how would you differentiate that

[stuck quartz]

power rule first?

[minor knot]

the first thing you gotta do is to write it down again

since the derivative of e^x is itself

so

[stuck quartz]

wait but

[minor knot]

ye?

[stuck quartz]

its not e^x in the problem

its e^(x^2+2x)?

[minor knot]

yea but

e^x is like the parent function

so you will be doing the same

write the thing again, but this time, you have to multiply it by the derivative of the exponent

[stuck quartz]

how is this

[minor knot]

yea thats right

[minor knot]

just put d/dx here so you dont forget

[stuck quartz]

Ok ty

[stuck quartz]

oh and the original example i gave you was wrong

instead of 2e^(1-x)^2

it was 2e^(1-x^2)

however i did what u told me

is this right

[minor knot]

lmao

yea thats right

[stuck quartz]

for 2e^(1-x^2)

Okay thank you

the only thing im still confused about is

the 2

[minor knot]

$e^{1-x^2}$

[potent lotus]

Renz

[stuck quartz]

why do we not do anything with the 2

[minor knot]

which 2

ohh

[stuck quartz]

the one thats being multipled to this

do we not use product rule or anything when solving for the derivative???

[minor knot]

its just a constant being multiplied

[stuck quartz]

cause i swear that constant would be 0

[minor knot]

so no need to worry about it

[stuck quartz]

derivative of 2 would be 0

though

[minor knot]

for example if you have 5x²

$\frac{d}{dx}5x^2$

[potent lotus]

Renz

[minor knot]

the 5 is a constant

[stuck quartz]

derivative of that would be 2x

[minor knot]

you can factor that out from the derivative operator

[polar frost]

10x

[minor knot]

$5(\frac{d}{dx}x^2)$

like that

[potent lotus]

Renz

[minor knot]

and yk the derivative of x² is just 2x

so it becomes

,,5(2x) = 10x

[potent lotus]

Renz

[minor knot]

the point is, the thing being multiplied was just a constant

you will only apply product rule if its a variable that is being differentiated

for example

,,(5x^2)(x+2)

[potent lotus]

Renz

[minor knot]

you have to apply product rule there

wait bad example

,,(e^x)(sinx)

[potent lotus]

Renz

[minor knot]

there you have to apply product rule, since both factors have x in it

[stuck quartz]

Oh

that makes sense

[minor knot]

,,(e^x)(sin\pi)

[potent lotus]

Renz

[minor knot]

but if you have smth like this

[stuck quartz]

i didnt know this was a thing 😭

ty so much

[minor knot]

you can just differtiate e^x and multiply sin(pi)

since sin(pi) is just a constant

yea

just remember, you can only factor out constants

that will be more common once you get to integrating stuff

[stuck quartz]

Okay bet ty so much again

.close

[calm coral]

Channel closed

Closed by @stuck quartz

Use .reopen if this was a mistake.

[minor knot]

yw

[calm coral]

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[stuck quartz]

hey im back again

can you help me find the derivative of this

did i do something wrong?

quotient rule i used

3^(2t)/t

[calm coral]

@stuck quartz Has your question been resolved?

Channel closed

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Remember:
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[hasty creek]

Are a and x the same thing here?

[hasty creek]

It seems to be approaching -1, and it says x = -1

what even is f(a)

[untold compass]

a is the point you're looking at for the limit, here it's -1

whereas x is a variable that can move around

[hasty creek]

so x and a are the same thing here?

[civic dirge]

no

[hasty creek]

it says x = -1

and it says we're approaching -1

[civic dirge]

You're ignoring a lot of that sentence

At the point x = -1, continuity is violated.

That doesn't mean x is now fixed to be -1 for all further occurrences of x

[hasty creek]

oh

f(-1)=-3, and the limit as x approaches -1 at least from the left side is 3

and as it approaches from the right side its -3

so is the limit non existant

[civic dirge]

Yep! That's true

[hasty creek]

ok awesome

.close

[calm coral]

Channel closed

Closed by @hasty creek

Use .reopen if this was a mistake.

[calm coral]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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

[candid pilot]

So i drew the reciprocal of the left equation

Is this correct?

[polar frost]

check this part too

[candid pilot]

There wasnt one for the question though

[polar frost]

Usually, we will use limit to find them

Base on the graph alone isnt really enough

[candid pilot]

Is the reciprocal of the graph on the right correct?

[polar frost]

Look like it

yeah

[candid pilot]

Isnt there supposed to be a third graph in the middle or no?