help-23

[feral linden]

My original thought is to let you use product rule directly, 1 times e times e

But anyway, correct

[lean otter]

but I'm confused, you see where I added +1 I also have -1 and only once did that -1 get lost where it disappeared

beginning

when i writte 1+ expresion-1

where -1 go?

[feral linden]

I didn’t add or minus anything

It’s just (n+3)/(n+2)=(n+2+1)/(n+2)=1+1/(n+2)

[lean otter]

oke

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[grim pendant]

yo

[grim pendant]

could i get some help with this question please

its diff calc btw

ive got to a point where its 4/(2+h) -2 all over h

but im stuck on that part

[desert pasture]

you aren't allowed to use the rules for diff., right?

[grim pendant]

im not to sure what you mean sorry we just started the topic

heres an example from my text book if you understand it better

[tidal imp]

You’re restricted to the limit definition of derivative probably

Yes

Ok so

Combine the numerator into one fraction

[desert pasture]

If not, the differential at an arbitrary point x would be $\frac{\frac{4}{x+h} -\frac{4}{x}}{h}$. not what you wrote(where h approachs zero)

[flat frigate]

physicsrocks

[grim pendant]

right

so you end up with (4/h)/h ?

[tidal imp]

Show your work

Because the goal is to cancel h out at some point to “patch the hole”

[grim pendant]

ok

but thats the correct direction right

having (4/h)/h

[desert pasture]

no, that's wrong, you'll get the differential is 4 then

[tidal imp]

I think you confused yourself

[grim pendant]

yeah im lost

[tidal imp]

You wrote 4/(2+h)-2 and then did 4/((2+h)-2)

When it is actually (4/(2+h))-2

[grim pendant]

oh

right

[tidal imp]

This is why you shouldn’t write fractions like this on paper

Use the horizontal line

[grim pendant]

i do

i just dont know how to use the horizontal line on the server

[tidal imp]

\frac{num}{den}

$ on both sides

$\lim_{h\rightarrow0}\frac{\frac{4}{2+h}-2}{h}$

[flat frigate]

CST

[tidal imp]

This is what you have to simplify, right?

[grim pendant]

yeah

[tidal imp]

Ok, so you want to get rid of complex fractions like this

[grim pendant]

and then that goes to $\frac{\frac{4}{2H+h^2}$

[flat frigate]

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

[grim pendant]

ok i think i get it

cheets

cheers

close.

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[raw osprey]

@serene mango

[safe radish]

Please don't occupy multiple help channels.

[serene mango]

tq

Let m be a fixed non-zero integer. For integer a,b, we say that they are congruent modulo m iff a-b is divisible by m. We write this as a
≡
b (mod m). Let R be the relation on the set Z of integers defined by aRb iff a
≡
b (mod m). Show that R is an equivalence relation on Z.

[raw osprey]

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[valid onyx]

So I have this physics equation $60t = 60-120(t + 0,25)$ with t being the time in hours. 60 and 120 being the speed in km/h , 0.25 being 1/4 of an hour and finally 60 km, now since the units in this equation aren't the standard units, I decided converting 60 to 16.67m/s, 120 to 33.33 m/s, 60 to 60 000 m, 0.25h to 900 s. This modified equation gives me 600 seconds which translates to 10 minutes and not 30, any idea why this modified equation is wrong?

[harsh parcel]

200 or 120?

in original eq its 200km/h

in your conversion you have converted 120

[flat frigate]

froud

[valid onyx]

yes sorry, 120 @harsh parcel

[harsh parcel]

im getting

10 minutes as the asnwer

not 30

60t=60-120t-30

180t=30

t=1/6hrs

10min

[thin bridge]

can you show the original question

[valid onyx]

yes thats the thing, I'm supposed to get 1/2h for the original equation

[thin bridge]

was this equation given to you?

[valid onyx]

this is the original equation with the result

and my equation gives 10 minutes

[harsh parcel]

theres a mistake

in the solution

180t=30

they also get this

so t is 1/6

[calm bridge]

if the answer is 1/2h, there should be t - 0.25 in the original equation.

[valid onyx]

oh I see, there's a mistake then

Here is the original problem, did I do a mistake in the equation?

Train A leaves city A to go to city B at 8:00 am with a constant speed of 60km/h, Train B leaves city B to go to city A at 8:15 am with a constant speed of 120km/h, given that the distance between both cities is 60 km at what time will the trains cross @calm bridge

[calm bridge]

answer is indeed 1/2h

[thin bridge]

yeh, you should've had t-0.25

[calm bridge]

but idk how to form the equation lol (i used a different approach)

[valid onyx]

ohhh right

Yeah that's the mistake, well I'll tell the teacher then

thanks everyone

[thin bridge]

B is leaving 15 minutes later thus is travelling for 15min less

[valid onyx]

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[primal onyx]

now i know that for the first one the answer is c

but my question is how do i get to that answer

[desert pasture]

What's the question? Match the following?

[desert pasture]

Do you know the laws of indices?

If so, apply them

[primal onyx]

so like (ab)² = a²b² that kind of stuff u mean

[desert pasture]

Yup

Apply those rules here

[primal onyx]

see what i did was (i multiplied all the exponents with the -5 but that gave me the completely wrong answer

[desert pasture]

Also remember $a^{-x}=\frac{1}{a^x}$

[flat frigate]

physicsrocks

dldh06

[desert pasture]

What command did you use to get this?

[calm bridge]

green role

[worthy hemlock]

Secret ones

[calm bridge]

really?

[flat frigate]

Dyssrupt

[calm bridge]

[worthy hemlock]

Because I'm good with texit

[worthy hemlock]

You know the secret too with texit

[primal onyx]

so i would push b down and then when doing the -5 i would bring it back up etc?

[desert pasture]

Yes

[primal onyx]

ahh okey thank you

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[mint radish]

Idk what the error is

[fluid token]

do you know how logs work?

[mint radish]

yeah

but i don’t see an error

?

[fluid token]

can you identify your base and exponent?

[mint radish]

So it should be log10(25)=t?

[fluid token]

yup

[mint radish]

so what does t equal then?

1.39794000867?

[fluid token]

this

[mint radish]

What is that as a fraction

Originally there is t=25/10

So do I write

25 = t/10?

[fluid token]

that’s the error..

[mint radish]

you just don’t do that?

[fluid token]

nope, no need

you can check your answer by using the value you got for t in the original equation

[mint radish]

My teacher prefers fractions as answers

[mint radish]

Wdym

[fluid token]

B = 100(10)^t

[mint radish]

so I write

10=100(10)^t?

@fluid token

[fluid token]

no, i’m saying you can check your answer by using the value you got for t

[mint radish]

how

[fluid token]

,w calc 100(10)^(1.39794000867)

[flat frigate]

Results provided by WolframAlpha

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

Requested by __saad

[fluid token]

like this

[mint radish]

ok ty

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[candid wolf]

Heyy guyss! I just need help real quick this wont take too long

[candid wolf]

how do you solve 2^10+1?

[hardy lion]

You have to work out 2^10 by hand

it might be easier to do (2^5)^2

There isnt really a good way to do exponents

[mortal thunder]

it's deceiving,

[candid wolf]

Soo like raise 2 to the power of 10 first? what happens to the 1?

[mortal thunder]

it could be (2^10)+1 or 2^(10+1)

[calm bridge]

then add 1 to the final result

[candid wolf]

So like 2^10 1024 plus one 1025?

[hard crest]

yep

remember PEMDAS (or BODMAS or BIDMAS or whatever you use)

exponents, then addition later

[candid wolf]

Ohhh okie soo i just raise it to what comes first right? Then add the rest later?

[hard crest]

yeah

[candid wolf]

Okieee thanks everyone!

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[dark cargo]

i dont get directed angles.

[dark cargo]

it says directed angle fkd = directed angle fbd. what happens if we take dbf instead of fbd? what is the reasoning behind taking fbd?

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@dark cargo Has your question been resolved?

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@dark cargo Has your question been resolved?

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@dark cargo Has your question been resolved?

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[gritty estuary]

How should I approach solving for x here? I've tried taking a common denominator like y=x(1/(1-(1/x))) but that doesn't work

[proper crypt]

Start by multiplying by (x - 1)

Then try to get all of the x terms on one side

[safe radish]

@gritty estuary Has your question been resolved?

[gritty estuary]

So I am a bit further down the line. Still trying to solve for x and I think my way of going forward from line 3 might be wrong because I feel like I'm stuck after line 5

Oops I think x/(x-1)=y multiplied by (x-1) turns into x=y(x-1) and not x=y*x-1 at the first line which would be my mistake here

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[tall glade]

I don’t know how to solve this question

Would it be C

Since the area of the circle minus the area of the square is 28.539?

<@&286206848099549185>

[plucky elk]

,rotate

[tall glade]

<@&286206848099549185>

[vital dirge]

yeah

dunno why you calculated it exactly though

[tall glade]

What do you mean by that

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[tall glade]

I don’t understand this question at all

[royal kiln]

!status

[safe radish]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[tall glade]

1

[sweet mica]

You don't even know your question?

You didn't ask

[worthy hemlock]

@sweet mica Pretty this was their question

They didn't properly open a help channel

[sweet mica]

Oh, thank you

OK, so what equals the shaded area?

In terms of the area of the triangle and the area of the circle sector?

[tall glade]

I honestly don’t know

It’s question 6

@sweet mica

<@&286206848099549185>

[sweet mica]

The are of the shadow region is the area if the rectangle minus the are of the circle sector

Do you understand that?

[tall glade]

Sorry no

[sweet mica]

The area of the triangle is the total triangle

if you substract the area of the circle sector the only area left is the shaded area

[tall glade]

What’s the area of the triangle

[sweet mica]

The area is the surface

[tall glade]

Is it 6

[sweet mica]

No, the base is 6, what is the height of the triangle?

[tall glade]

Isn’t it 1/2(6)(h)

[sweet mica]

I'm asking what is the value of h

[tall glade]

10?

[sweet mica]

You know the base 6 and the angle θ

No

It depends on the angle

It can't be a fixed number

[tall glade]

6tan(theta)

[sweet mica]

Well done

So what is the area of the triangle

[tall glade]

1/2(6)(6tan0)

[sweet mica]

18tan(θ)

Good

Now, what is the area of the circular sector of radious 6 and angle θ?

[tall glade]

I don’t understand

1/2(6)^2(0)?

Wait wouldn’t the area be 18tan0

And then the area of the sector is 180

So it’s 18(tan0-0)

@sweet mica

[sweet mica]

Yes

Area of triangle: 18tan(θ)
Area of sector: 18θ
Area of shaded region: 18(tan(θ)-θ)

[tall glade]

Ok thank you

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[crimson idol]

i’m confused about what the inverse would do, also are 1 and 2 correct?

[crimson idol]

,rotate

[agile phoenix]

c is 4/15

d is 5/4

[crimson idol]

how does the notation change between c and d rn

like i’m confused on how putting the -1 in different spot changes it

like how does it change it

[thin bridge]

in b), the whole expression is being raised to the power of -1
in c), the inverse function is being notated
in d) only p is being raised to the power of -1

[crimson idol]

what does the inverse function being notated mean

[thin bridge]

f^-1 represents the inverse function of f

[crimson idol]

ok so the inverse would be x/5 + 1/5

i think i got it

thanks

i have another question tho

[thin bridge]

f^-1(x) = x/5 + 1/5
would work, yes

[thin bridge]

though a more efficient way would be
applying the definition of the inverse
f(f^-1(p)) = f(3)
p = f(3)

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@crimson idol Has your question been resolved?

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@crimson idol Has your question been resolved?

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[lean otter]

this is the quadratic formula i'm doing here

I see that it is a complex number

how to write

[hidden anvil]

i = sqrt(-1)

[lean otter]

can you draw the procedure for me or write it?

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[acoustic sentinel]

Evaluate the limit using the squeeze theorem. Use a calculator to graph f(x)
, g(x)
, and h(x)
when possible.

[acoustic sentinel]

if I plug in 0.0001 and -0.0001

they both give me infinity

so lim g(x) theta->0 has to equal infinity

Am I getting smth wrong here

[royal horizon]

That's werid they should give you a value near 0

[acoustic sentinel]

why

I mean multiply an infitiely small number by an infinitely large number that's not as large

[royal horizon]

Don't forget about θ² term

Why's cos(1/θ) large when θ is near 0?

Maybe you should try the squeeze theorem as mentioned in the question

It'll probably be simpler

[acoustic sentinel]

That's what I did my guy

[acoustic sentinel]

1/cos theta is large cuz 1/ a very small number is a very large number

1/.000001 is 1000000

Ah shit

I forgor

the cos

[royal horizon]

So you tried the squeeze theorem and then tried plugging in values but got different answers?

[acoustic sentinel]

nope, got the same answers

5.29 * 10^-12 or some shit

[royal horizon]

Can you show how you got that?

[acoustic sentinel]

plugged in 0.0001^2 * cos(1/0.0001)

into my calc

then I tried it with -0.0001

[royal horizon]

No, I mean the squeeze theorem way

[acoustic sentinel]

I did that and it gave me 0

on both sides

[royal horizon]

Ok that's normal

[acoustic sentinel]

ye but I felt like it was wrong

it looked weird

[royal horizon]

In what way?

(I thought the name was werid when I first knew about it)

[acoustic sentinel]

idk man it just looked sus

like it was wrong

squeeze theorum looks like it's made by someone who doesn't know shit about math

idek how it works tbh

it just looks weird af

[royal horizon]

So anyways, what did you use for g(x) and h(x) then?

[acoustic sentinel]

X^2 and -X^2

was I wrong?

[royal horizon]

Ok

No, it's right

[acoustic sentinel]

Ok, that's it, good night

[royal horizon]

Bye

[acoustic sentinel]

.close

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[open mantle]

What does it mean if there is a number line problem with a number four on the bottom in the fives place?

[lean otter]

divide

[open mantle]

What am I dividing in said problem?

The whole thing is throwing me off cause I'm used to the basic number line from middle school.

[lost patrol]

Yeah so basic things you need to know is, you have like, a scale.

1 place is equal to how many units

What you're used to is 1 place equal to 1 unit yk

[open mantle]

Well yes that's the basic basic one. I do know that sometimes it can be multiplying too

Such as maybe the 4 times family

[lost patrol]

Now...

[open mantle]

Hmmm

[lost patrol]

You don't have this

You have a bit different line

[open mantle]

Yeah I'm lost tbh

Yes

[lost patrol]

Each place has a constant increment you see

[lost patrol]

Like here, every next point on the number line is 1 more the previous one

[open mantle]

Hmmm

[lost patrol]

So similarly here

All those five places

They have a constant difference

Like, if it was multiples of 4 number line

The difference would be 4

0, 4, 8, 12, 16, 20 and so on

Here...

You have something like

0, i, ii, iii, iv, 4

where i, ii, iii and iv are values you don't know

But they all equally apart.

So do you get something out of it

[open mantle]

Are they whole numbers still?

[lost patrol]

No, not necessarily

And in this case they aren't

[open mantle]

Hmm that was my theory

[lost patrol]

Nope, a real number line contains all real numbers

Including the weird ones such as pi

[open mantle]

I think I'm following

[lost patrol]

yeah

Now do you have like a rough roadmap of how you'd wanna solve this

Now that you know it's not necessary the points marked on the line be real numbers

[open mantle]

I'm gonna look back over what you said real quick and get back to ya

[lost patrol]

Suree.

Do you know some algebra?

[open mantle]

Ha some, I never fisished lol. This is actually a problem a friend is working on. It bothered me I couldn't figure it out

[lost patrol]

Ah haha

Can you form basic algebraic equations and solve them?

[open mantle]

I think so? lol

[lost patrol]

Haha

So

[open mantle]

Like finding x

[lost patrol]

See, everytime you move to a next point

The value increases by k

And after jumping five points after 0

The value is 4

If you equally jumped on each point

How much did you

jump

[open mantle]

So it's a fraction of a number each jump?

[lost patrol]

It can be anything haha, here it is a fraction yes

[open mantle]

Ok that makes sense

[lost patrol]

So

You make five equal jumps

You start from 0 but land at 4

So how much did you cover in each jump

[open mantle]

Bro my brain hurts lol, I'm close though

[lost patrol]

Haha it's okay

Take it easy

I'll give you a different question

See.

You have to cover a distance of 10m in 2 jumps

How much distance do you need to cover in 1 jump

[open mantle]

Each jump is worth 5m?

[lost patrol]

Yeah

How did you get this answer?

Like, what did you do to get this answer

[open mantle]

I don't exactly know lol

Something knocking around from algebra

Solving for x basically

[lost patrol]

No

What did you basically do

you had to cover 10 m in 2 jumps

You wanted to tell me how much would u cover in 1 jump

Unitary method basically

Innit?

[open mantle]

I suppose

I honestly don't remember a ton of math

[lost patrol]

I'll tell you

You divided 10m in 2

10/2 = 5 m

Which gave you what you wanted

Easy innit?

[open mantle]

True

[lost patrol]

Similarly

You make five jumps starting from zero

To cover 4m

How much is each jump

[open mantle]

Hmm I'm so close, I get the concept

Sorry I feel dumb for not getting this by now

[lost patrol]

It's okay

I won't give you the answer

But you'd divide 4m equally into 5 jumps

[open mantle]

I honestly tried that but fractions have always messed with me

[lost patrol]

Leave it in a fraction

You can do the division later

You need to find the third place

[open mantle]

What is the third place?

My brain is mushing at this point

Well thanks for the help, I think I get it a bit better now

. closed

Wrong one lol

.close

[safe radish]

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[quiet meteor]

I keep getting many answers for R

[rancid sand]

Pythgoras?

[quiet meteor]

?

[rancid sand]

Now ur finding the length of R right?

[quiet meteor]

Yes if hopefully 01 and 02 are correct

Can u check

[rancid sand]

with a smaller mistake ig

that's technically not tan^-1

that's arctan

[quiet meteor]

But don’t you have to do -1 to find sides

[burnt notch]

A lot of people and books and also calculators write that instead of arctan

[rancid sand]

tan^-1 is not equal to arctan techniqcally

[rancid sand]

ys ik

[quiet meteor]

Sooooo

[burnt notch]

By convention it is

[burnt notch]

The angles you got are correct

[rancid sand]

use cos rule

[quiet meteor]

So I am correct for 01 and 02?

[rancid sand]

ys

[quiet meteor]

Oh ok

[burnt notch]

Now do Pythagoras theorem to find the hypothenuse R

[quiet meteor]

OH

right

I totally forgot

[rancid sand]

HUH

The angle is not 90 degree

[quiet meteor]

Uh

[rancid sand]

33.7 + 56.31

is 90.01

Not 90

U can't use pyth

Instead u need to use cosine rule

which $$ c^2 = a^2 + b^2 - 2ab(cosC)$$

[flat frigate]

Fossil

[quiet meteor]

I am going to be honest with you I never did that for a side in geo

[rancid sand]

lol

[quiet meteor]

So is that a different method

[rancid sand]

ye that's how I do if me

[quiet meteor]

Because I don’t think that’s what I should do

[rancid sand]

lol

what ur teacher tech u

teach

[quiet meteor]

Honestly idk even know at this point 💀

[rancid sand]

Wait

[quiet meteor]

I am going into physics

[rancid sand]

it's 90

[quiet meteor]

Yea that would make sence

[rancid sand]

NO

[quiet meteor]

Bruh

[rancid sand]

02 is incorrect

[quiet meteor]

How

[rancid sand]

finally, lol

[quiet meteor]

Let me do it again

[rancid sand]

the angle should be 33.69006753

which is 33.70

[quiet meteor]

One sec

[rancid sand]

ye

done

use Pyth now

[quiet meteor]

Alr one sec

Wait why does it have to be 33.70 and not 33.7

Wait I am confused now

It’s both 90.1

???

[rancid sand]

HUH

[rancid sand]

Correct to 2 dp

[quiet meteor]

?

So 33.70

[rancid sand]

ys

[quiet meteor]

But it still adds up to 90.1

[rancid sand]

NO

It didn't

[quiet meteor]

I am doing it rn tho

[rancid sand]

33.70 + 56.30

[lilac patio]

did you round wrong

[rancid sand]

wait

Oh oh

Someth'in big happens

[quiet meteor]

It’s 56.31

[rancid sand]

my calcualtor can add number with irrtional number

the correct thing is 56.30993247

[lilac patio]

,w compute arctan(20/30)

[flat frigate]

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[lilac patio]

,w compute arctan(30/20)

[flat frigate]

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Requested by Arctic#1647

[lilac patio]

ughhh radians

[quiet meteor]

What’s happening

[quiet meteor]

What

[lilac patio]

,w compute (0.588+0.982794)*180/pi)

[flat frigate]

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[lilac patio]

it adds up to 90

you rounded off to like the first or second decimal

[quiet meteor]

Oh wait

tan-1(30/20) is 56.3 right

[lilac patio]

more decimals

[quiet meteor]

?

[lilac patio]

,w compute arctan(30/20)*180/pi

[flat frigate]

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Requested by Arctic#1647

[lilac patio]

ok 56.31 is good

[quiet meteor]

Yea it is 56.3 not 56.31 right?

But then it makes 90.1

[lilac patio]

,w compute arctan(20/30)*180/pi

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Arctic#1647

[lilac patio]

33.69

33.69+56.31

[quiet meteor]

So it has to be the same amount of numbers for each for it to work?

[lilac patio]

yes

[quiet meteor]

Ohhhhhhh

Wait but can’t 56.3 also workout

[lilac patio]

if you do one value yeah

but this isnt really the important stuff

[quiet meteor]

So I can do that?

[lilac patio]

just round correctly

sure

[quiet meteor]

So rounding correctly would be which one

[lilac patio]

im not even sure where this issue was brought up

[lilac patio]

either one

but it has to be consistent

[quiet meteor]

Right so it has to be for both 01 and 02 has to be same amount of numbers for it to work

Ok I am just going to do 56.3 + 33.7

That what I would normally do

That is correct right

@lilac patio

[lilac patio]

yep

[quiet meteor]

Wow this was a lot for a simple thing for R

Alr thanks

.close

[safe radish]

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[astral lodge]

Hello, can someone help me with the equatios with fractions?

[astral lodge]

I don't really know how to start

[lilac patio]

you want to get both sides to have the same denominator

you would do that by multiplying a certain fraction by a special 1

a fraction that equals 1 but changes the denominator

like if you wanted the left side to be one denominator

just multiply (x+5)/3 by 2/2

and they will have the same denominator

[astral lodge]

So, it will become - X + 10 / 6 - X + 15 / 6 = 1/3 - X + 3 / 4

that is X+20-X/6 = 1/3 - x + 3/4

20/6 = 1/3 - x + 3/4, X cancels, right?

[safe radish]

@astral lodge Has your question been resolved?

[safe radish]

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

Is my method correct

[lean otter]

Peace be upon u btw

[quasi bison]

ok so you found the area of a single slab and found how many slabs it'd take by area to fill the patio

that's a step in the right direction, however this assumes that the slabs can be broken into parts if needed.

[lean otter]

But

[lean otter]

Wdym

[quasi bison]

let me try to make an example...

if the patio was instead a 3.3m by 3.3m square

doing the same calculations you would find that it takes 30.25 tiles to cover

[lean otter]

Yea

[quasi bison]

this is what'd happen if you actually started laying the tiles down

those red bits are fractions of a full tile

[lean otter]

Oh

[quasi bison]

note that in your case that actually doesn't happen! since your patio is covered by a perfectly fitting 5×6 grid of tiles.

but still.

area alone couldn't tell you about these shenanigans.

[lean otter]

loool shenanigans hahaha XDD

[quasi bison]

you could have something like 8 by 5.5 or something where the number of tiles is whole by area but you still need to break them

[lean otter]

I see ur point

So u saying my method only works for perfect grid of tiles

@quasi bison

[quasi bison]

yes

it is dodgy in this way

it would be better to find how many tiles fit along each side of the patio first

then multiply that

if you get a non-integer on either the width or length you know something sus may be going on

[lean otter]

Like a decimal

Lol sus

@quasi bison

I got a decimal

Byyour meathod

[quasi bison]

is this for a different problem...?

[lean otter]

No

I madeir

It

[quasi bison]

yes but you made a new problem, different than the one you started with

and... you're now tiling an 8cm by 5.5cm rectangle with 4cm by 4cm tiles?

[lean otter]

i think if its not a perfect grid, u will get a non integer by either method

So i think its ok

(Mymethod)

@quasi bison

[quasi bison]

...

[lean otter]

Yes?

[lean otter]

What

[quasi bison]

i guess i'll say yes now to get you off my back

[lean otter]

This is what I got from your message

[quasi bison]

AI?

[lean otter]

lol

If my understanding is right (of ur mssgs), then thst mewns my method is okay

Which means ill close this wns get off from your back

.close

[safe radish]

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[gentle oak]

Find all integers $n$, such that :
$$\prod_{k=0}^{n-1}(2k+1) +\sum_{k=1}^{n-1}(2k-1)=t^2-1$$
With $t\in\mathbb{Z}$

[thick otter]

did you mean k=0 at the bottom of the product sign?

[flat frigate]

Joseph.P

[gentle oak]

Ye sorry

[thick otter]

Is t fixed?

[gentle oak]

It’s an integer

[thick otter]

Well this function is increasing in n so should there be a value, it is unique

[gentle oak]

Find all integers $n$, such that :
$$\prod_{k=0}^{n-1}(2k+1) +\sum_{k=1}^{n-1}(2k-1)=t^2-1$$
With $t\in\mathbb{Z}$

[flat frigate]

Joseph.P

[gentle oak]

It’s equal to $\dfrac{(2n)!}{2^n\cdot n!}+(n-1)^2$

[flat frigate]

Joseph.P

[gentle oak]

Can someone tell me if this is right ?

<@&286206848099549185>

[brave atlas]

Hope this helps

[gentle oak]

It’s the same thing

[brave atlas]

U had to open the bracket

Using (a-b)^2

Which is then

a^2 -2ab+b^2

[gentle oak]

a^2-2ab+b^2

I know

[brave atlas]

….

What does “it’s equal to mean”?

Is something equal to this or what….

[gentle oak]

$$\prod_{k=0}^{n-1}(2k+1) +\sum_{k=1}^{n}(2k-1)=\dfrac{(2n)!}{2^n\cdot n!}+n^2+1=t^2$$
With $t\in\mathbb{Z}$

[flat frigate]

Joseph.P

[gentle oak]

I changed n-1 with n for the sum

[gentle oak]

For the second line it’s $a_n=n^2+n$

[flat frigate]

Joseph.P

[gentle oak]

The twos simplify each other

[brave atlas]

Yea

Forgot about the 2’s

Yea just chop them off

[gentle oak]

But I don’t understand what your doing with $$\lim_{h\to+\infty} \sqrt{a_n}-\sqrt{b_n}$$

[flat frigate]

Joseph.P

[harsh parcel]

wait

[brave atlas]

Placing the values

[harsh parcel]

i think you tn is wrong

because isnt sigma 2k-1 =n^2

and not

what u got

i get answer as 3

but its bashing

[brave atlas]

Where…..

Line… which one

[harsh parcel]

.

[brave atlas]

1st or third line?

[gentle oak]

Ye that’s one wrong

[harsh parcel]

should be the factorial term +n^2+1

i put

2n!/2^nn! = 2n

and got answer as 3

[brave atlas]

Wait

[gentle oak]

I’m trying to create an exercise so the statement may change

[harsh parcel]

from observation we can definitely say that n has to be small

[brave atlas]

Lemme google it….if I get smth

[harsh parcel]

as rate of increase of factorial is much higher than power of 2

[gentle oak]

So for now the statement is
Find all integers $n$, such that :
$$\prod_{k=0}^{n-1}(2k+1) +\sum_{k=1}^{n}(2k-1)=t^2-1$$
With $t\in\mathbb{Z}$

[flat frigate]

Joseph.P

[brave atlas]

Bruh what did I find

[gentle oak]

I think it’s pretty wrong

[harsh parcel]

getting 3 for this

[brave atlas]

Yea,same

[gentle oak]

But how do you prove that’s the only solution

[brave atlas]

Try putting around 3 in places

Maybe then u can prove my answer wrong….I can be wrong

[gentle oak]

You mean 1/2 ?

If so it’s not possible because n is an integer

[harsh parcel]

my reasoning is

n has to be small

as factorial increases more

than 2^n

and then putting a few value at 4 it suddenly becomes very much more than 2n

so after that it can never reduce

however yes its not proper

its reasoning more so than a proof

[gentle oak]

Could you help me make it more complex

[harsh parcel]

put it =2n

ig from there showing it is easier

[gentle oak]

I don’t understand

[harsh parcel]

if this is a square

middle term has to be 2n

[gentle oak]

So (n+1)^2=t^2

[harsh parcel]

yeah

so putting the middle term =2n

but solving that in itself lol

i dont know if u could prove that

without saying what i did

because 2n! will increase much much more

it will be impossible for that to be 2n for larger numbers

huh lol

3 does satisfy

but

it gives 5^2

and not 4^2 as it should

so putting equal to 2n is wrong

[gentle oak]

And there’s an infinite amount of solutions for (n+1)^2=t^2

[harsh parcel]

not with the condition u have

[gentle oak]

I’m reposting the question.
Find all integers $n$, such that :
$$\prod_{k=0}^{n-1}(2k+1) +\sum_{k=1}^{n}(2k-1)=t^2-1$$
With $t\in\mathbb{Z}$

[flat frigate]

Joseph.P

[gentle oak]

We got 3 as an answer but we need to show that it’s the only one

[harsh parcel]

and if theres more to find that

[gentle oak]

<@&286206848099549185>

[safe radish]

@gentle oak Has your question been resolved?

[gentle oak]

Can somebody help me ?

[safe radish]

@gentle oak Has your question been resolved?

[gentle oak]

<@&286206848099549185>

[oak perch]

What

[gentle oak]

?

[lean otter]

U can join another maths server its has the same name and they reply fast

[gentle oak]

What’s the name

[lean otter]

“Mathematics” @gentle oak

[gentle oak]

Can you send me the link ?

[lean otter]

https://discord.gg/maths

[lean otter]

^

[gentle oak]

Thanks

[granite flower]

If you don't have more questions @gentle oak
Don't forget to close the channel using
.close to give the chance to others to be able to ask

[gentle oak]

My question is still not answered

[granite flower]

Oh sorry

[gentle oak]

No worries

[safe radish]

@gentle oak Has your question been resolved?

[gentle oak]

Here’s the question

[safe radish]

@gentle oak Has your question been resolved?

[broken yew]

worth trying in #elementary-number-theory imo

[gentle oak]

<@&286206848099549185>

Ok

[acoustic sentinel]

What is the limit

I tried conjugates but no luck

there

[gentle oak]

!help

[safe radish]

Please read #❓how-to-get-help

[acoustic sentinel]

Ik I just messed up, got it?

[white umbra]

Please don't advertise other servers here

[white umbra]

This is not your help channel, please reserve your own

[acoustic sentinel]

Brodie, if you took the time to read the messages you would know that it was a mistake, I forgot. That's it nothing more to it

[white umbra]

I read your messages, please be polite. It wasn't clear that you weren't referring to messing up on the problem rather than messing up in posting in someone else's channel. Don't take it personally.

[acoustic sentinel]

Bro, I never took it personally I was just clearing things up

[karmic hedge]

Hint: a number is a difference of squares iff it isn’t 2 (mod 4), so find when it is 2 (mod 4) and see which n don’t satisfy

[gentle oak]

It’s congruent to 2 mod 4 if n is pair and can’t be divided by 4

[karmic hedge]

I’ll check when I have time

[safe radish]

@gentle oak Has your question been resolved?

[safe radish]

Channel closed

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[fresh birch]

.reopen

ay

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
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[tame elk]

idk twtf

to do

i keep gettting 6

[quasi bison]

are you using a calculator?

[tame elk]

yeah

and like

myself

[quasi bison]

are you sure it is not set to degree mode?

[tame elk]

its not rlly hard

its just

.06/.01

[safe radish]

@tame elk Has your question been resolved?

[tame elk]

no

[safe radish]

@tame elk Has your question been resolved?

[tame elk]

LITERALLY NOOOO

[rigid inlet]

,w compute (sin(pi/2 + 0.1) - sin(pi/2))/0.1

[flat frigate]

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[rigid inlet]

@tame elk Can I see what you're putting into your calculator?

Whoops missed a 6, whatever

So roughly -0.3

[safe radish]

@tame elk Has your question been resolved?

[safe radish]

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[lean otter]

Can anyone help me with this?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[cold aurora]

What have you tried?

Or do you need help getting started?

[lean otter]

Hey! well I've tried typing it into this online integral engine

[cold aurora]

Lol. Did that not work?

,w integrate sin(ax) * cosh(x) * dx

[lean otter]

it did but I'm so confused

[flat frigate]

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[cold aurora]

Okay.

[lean otter]

well first of all we are probably using substitution right?

[cold aurora]

So, if you have no idea where to start - you need to use integration by parts.

[cold aurora]

Umm... Not really needed. We, sure, could use substitution to make life a tiny bit easy but it's not really needed.

[lean otter]

so I try to integrate sin(ax) first but it turns out the integration of sin(ax) is -cos(ax)/a which is completely confusing to me, 1.) why can we just treat a as a constant when its inside the brackets of a trigonomic function and 2.) if we are dividing by that constant why aren't we also taking x to the power of 2?

i didnt mean subsitution sorry i meant partially integrating i misspoke sorry

[cold aurora]

Well, integration of sin(ax) is -cos(ax)/a + c.
I dunno why are you confused by that.

[lean otter]

sorry if my math lingo is not on point I'm austrian

[cold aurora]

Also, in your first question - did you mean "can't" ?

Btw, we aren't treating a as constant. It is a constant. In fact, in an integrand, every other variable than the variable of integration is constant.

[lean otter]

ok lets put it this way: if I wanted to integrate 2x it would turn into 1/2x² right? so why if i integrate ax soesnt it turn into 1/ax^a?

[cold aurora]

No.

$\int 2x \dd{x} = 2\cdot \frac{x^2}{2}+C$

[flat frigate]

Enemagneto

[cold aurora]

That is just $x^2 + C$

[flat frigate]

Enemagneto

[cold aurora]

2 in 2x is a constant so it can be taken out of integration.

Basically

$\int 2x \dd{x} =2\cdot \int x \dd{x} = 2\cdot \left(\frac{x^2}{2}+C\right) = x^2 + C$

[flat frigate]

Enemagneto

[cold aurora]

Now, ask if any doubts still.

[lean otter]

oh yeah i get that

let me think a bit

so if we used a instead of 2 what would that look like?

like ax dx

$\int ax \dd{x} =2\cdot \int x \dd{x} = a\cdot \left(\frac{x^2}{a}+C\right) = x^2 + C$

not like this right?

[cold aurora]

No.

[flat frigate]

Looter

[lean otter]

shit

well I'm confused lol

[cold aurora]

Still wrong.

$\int x^n \dd{x} = \frac{x^{n+1}}{n+1} + C$

[flat frigate]

Enemagneto

[cold aurora]

When we had $x$, it was basically $x^1$.

[flat frigate]

Enemagneto

[cold aurora]

So, it became $\frac{x^{1+1}}{1+1} + C$

[flat frigate]

Enemagneto

[lean otter]

yeah exactly so shouldnt the result be a*(x^2/2)

[cold aurora]

It would be that only.

Who said it wouldn't be?

Of course, there should be a +C in there.

[lean otter]

because in the original problem it said:
f(x)=sin(ax)
F(x)=-cos(ax)/a

[cold aurora]

Yes. Because that's sine function.

[lean otter]

oh so there is a special rule im not aware of?

[cold aurora]

Integration of sine function is different from general monomial.

[lean otter]

ohhh ok

[cold aurora]

Well, there are some major rules.

However, integration of some of the common trigonometric functions is often expected to be remembered as a basic.

For example:

[lean otter]

nono I get the whole sin(x)->cos(x)->-sin(x)->-cos(x)->sin(x)

[cold aurora]

$$\int \sin{x} \dd{x} = -\cos{x} + c$$
$$\int \cos{x} \dd{x} = \sin{x} + c$$

[flat frigate]

Enemagneto

[lean otter]

but the a is just really confusing to me

[cold aurora]

Okay. See. That's an issue which can be dealt with using a u-substitution.

Let's try doing it.

We have $\int \sin(ax) \dd{x}$.

[flat frigate]

Enemagneto

[cold aurora]

Let's assume $ax$ to be $u$.

[flat frigate]

Enemagneto

[cold aurora]

Good so far?

[lean otter]

yes

[cold aurora]

Just to be clear, we are assuming ax to be u so that we get a term of sin(u) which we know how to integrate. We might get some constant outside which can just simply be taken outside of the integrand.

Remember that you can't do that (taking variable out of the integrand) when constant is in argument of some trigonometric/logarithm/exponential or such function.

Anyway, so $u = ax$

[flat frigate]

Enemagneto

[cold aurora]

Now, when we differentiate with respect to $x$.

[flat frigate]

Enemagneto

[cold aurora]

$\frac{\dd{u}}{\dd{x}} = \frac{\dd {(ax)}}{\dd{x}} = a\cdot \frac{\dd {(x)}}{\dd{x}} = a$

[flat frigate]

Enemagneto

[cold aurora]

Fixed

Makes sense?

[lean otter]

oh you used the substituion to explain to me whats going on

not to solve the equation

got it

[cold aurora]

So, do i continue this?

[lean otter]

that makes so much sense now

Thank you

like a lot

so it's a good technique in general to do this when I can't figure it out

[cold aurora]

,tex
\text{Now, we just substitute.}
$\int \sin(ax) \dd{x} &= \int \sin(u) \frac{\dd{u}}{a} &= $\int \frac{sin(u)}{a} \dd{u} &= a\cdot $\int \sin(u) \dd{u}$

[flat frigate]

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

[cold aurora]

Basically that. Now, you know how to integrate sin(u).

As that is simply -cos(u) + c

Now, try your original question.

[lean otter]

Ok I will but it could take a bit 😅

[safe radish]

@lean otter Has your question been resolved?

Channel closed

Closed by @green dirge

Use .reopen if this was a mistake.

[safe radish]

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[barren vale]

guys how to solve without u substitution

[thin bridge]

why are you trying to do this without sub

[barren vale]

because our teacher wants us to

he makes us hide the u sub

and do it in our head

[thin bridge]

you'd essentially still require applying chain rules,
the difference is whether you do it clearly on paper
or try imagining it in your head and potentially make mistakes

[barren vale]

yeah

thats what he makes us do

ik the rules

[thin bridge]

so if you have trouble visualising it in your head, do it with sub

[barren vale]

but i cant

[thin bridge]

until you get comfortable enough to do it in your head

[barren vale]

i can do most problems in my head

[thin bridge]

hence why you should do it with sub first, it takes experience and practice

[barren vale]

yeah but i cant do it with sub

or else i would have used it

the teacher doesnt let us

[thin bridge]

not allowed to or unable to do it even if you were allowed

[barren vale]

not allowdd

[thin bridge]

so IF allowed to do it with sub, you'd be able to do this?

[barren vale]

yes

[thin bridge]

have you actually done this with sub yet?

I don't care what the teacher says

we can't really help you with mental visualisation like this

[barren vale]

one sec

[thin bridge]

with an explicit sub, it makes it easier to see how chain rule is relevant here

[dull sequoia]

What you might notice is this form $\int g’(x)f(g(x)), dx$

[flat frigate]

Frosst

[thin bridge]

which makes it easier to think about and visualise what's happening in future examples

[dull sequoia]

Which looks strikingly like the chain rule

[thin bridge]

don't forget the sign

[dull sequoia]

Yes the signs are important

[thin bridge]

also this is technically a 2-level chain rule problem

[barren vale]

you sub the fraction

sorry i got a call

[thin bridge]

so yeh my main point is

so if you have trouble visualising it in your head, do it with sub
until you get comfortable enough to do it in your head

[barren vale]

yeah i need to know how to do it with inverse chain rule