help-26

[glass badger]

(1-a)^2 - (5-a)^2 + (6-b)^2 - (6-b)^2 = c^2 - c^2

It would be like this

How many unknowns are left?

[spring ivy]

2

[glass badger]

Which are?

[spring ivy]

the first two terms?

[glass badger]

Yep، you have only a, right?

[spring ivy]

yeah

[glass badger]

So you can solve for it now

Solve for a and tell me what it is

[spring ivy]

-4-a^2?

[glass badger]

(1-a)^2-(5-a)^2=0

[spring ivy]

what

how

how is it 0

ohh

wait

[glass badger]

Both sides of the equation ate c^2

We subtracted them, remember?

You only need to solve for a now. Expand both (1-a)^2 and -(5-a)^2, collect the terms and then solve for a

[spring ivy]

like this?

[glass badger]

This is incorrect

This is a systems of equations with no possible solution

[spring ivy]

I’m so bad at this

[glass badger]

Don't worry, you'll get better

Do you know how to square a binomial?

[spring ivy]

I can’t even figure out a problem

[spring ivy]

Show example

[glass badger]

Like (a+b)^2

Expanding it out

[spring ivy]

Not really

[glass badger]

a^2+2ab+b^2

You should work on on your algebra, stuff like binomial and the like are essential

[spring ivy]

ok

[glass badger]

Try applying the above formula to 1-a and 5-a

[spring ivy]

idk bro

i’m not gonna figure this out

this is so confusing

[glass badger]

alright, hold on. I can simplify it further

[spring ivy]

can you just give me the answers please. I will never figure this out and I have homework checks so if I haven’t done it I will get a 0

[glass badger]

I'm sorry, I can't just give you all the answers as that's discouraged.

[spring ivy]

This is impossible 😓

[glass badger]

Do it slowly. You know systems of equations so I reckon you're good enough to understand this

[spring ivy]

I don’t know how to do it though

[glass badger]

Okay, for expanding (1-a)^2, try thinking of it as (1-a)(1-a)

[spring ivy]

yeah

[glass badger]

Do you know how to multiple two expressions like this?

[spring ivy]

um

like multiply?

[glass badger]

Ye

[spring ivy]

like 1•1. 1•-a -a•1 -a•-a

?

[glass badger]

Yep exactly

[spring ivy]

oh

ok

yeah i can do that

[glass badger]

See? You already know how to do it. My fault for complicating it with all the unnecessary equations.

[spring ivy]

Ok

It’s ok

[glass badger]

You can now solve (1-a)^2-(5-a)^2=0

[spring ivy]

like that?

[glass badger]

Yup, just one thing. Don't forget to change the signs. You're subtracting (25 -10a +a^2) so what happens to the signs?

[spring ivy]

i have to go to sleep in 10
minutes and i only answered 2 questions when I’ve been doing this for like 2 hours😕

[spring ivy]

oh yeah you flip?

[glass badger]

Yep

When do you have to hand in this this problem sheet?

[spring ivy]

Tomorrow

First period

So like that?

[glass badger]

Yep, now you can very easily solve for a

[spring ivy]

a=3

[glass badger]

Yep

[spring ivy]

then what?

[glass badger]

Now to solve for b, you use the same process

[spring ivy]

but where is the b term

[glass badger]

Go back to the 3 equations you wrote earlier

[spring ivy]

here

[glass badger]

Yep

[spring ivy]

ok

then

[glass badger]

What do you do?

[spring ivy]

plug in 3 for a

[glass badger]

You can, and that's not wrong perse. but there is a way to get rid of all the a's like you did with the b's before

Look at the second and third equations

[spring ivy]

ok

[glass badger]

Do they both have (5-a)^2?

[spring ivy]

you can eliminate the (5-a)^2

[glass badger]

Yep

[spring ivy]

yeah

[glass badger]

You can also eliminate c^2

And b becomes the only unknown variable

Try solving for b now

[spring ivy]

Ok

like this?

i think i messed up

[glass badger]

Ywp, watch out for the signs

[spring ivy]

it’s supposed to be 36-12b=0 then -12b=-36 then b =3

[glass badger]

Yes

Correct

[spring ivy]

Yay

[glass badger]

You found a and b

[spring ivy]

so what is C

[glass badger]

How do you find c now?

[spring ivy]

hm

idk

lol

[glass badger]

You can use any of the 3 equations

[spring ivy]

ok let me try

[glass badger]

You know both a and b

[spring ivy]

if i use the B equation

(5-3)^2 + (6-3)^2=c2

i’m confused

idk how

[glass badger]

You did it correctly

You have only one unknown now which c

[spring ivy]

but idk what to do

[glass badger]

Whats 5-3?

2

[spring ivy]

2

[glass badger]

Square it

[spring ivy]

then 2^2 is 4

[glass badger]

4

Do the same for the other

6-3

[spring ivy]

6-3=3 3^2=9 4+9=15

[glass badger]

Then square it

[spring ivy]

15=c^2

[glass badger]

Yep

So c= sqrt(15)

And there you have it

[spring ivy]

ok

[glass badger]

Plug a, b and c into the general formula of the circle

[spring ivy]

Wait what

[glass badger]

And you have your equation

[spring ivy]

how

[glass badger]

(x-a)^2+(x-b)^2=c^2

[spring ivy]

oh

there

now what

[glass badger]

Correct for the most part. One little mistake though, c

[spring ivy]

?

would it be 225?

or just c^2

[glass badger]

c^2=15, so c= sqrt 15

[spring ivy]

oh

[glass badger]

You wrote c^8

[spring ivy]

wdym

[glass badger]

C=sqrt(15)
C^2=15
C^3=15sqrt(15)
.
.
.
C^8=15^2

[spring ivy]

so what would i put

instead of 15^2

[glass badger]

Put c^2, which is 15

[spring ivy]

ok

[glass badger]

You can use this same method for all the other three point questions. It's a bit of a doozy but not very complicated one you get the gist of it

Now you can graph it if you want

[spring ivy]

How would you do that

[glass badger]

I told you before

In thus picture

[spring ivy]

ok

i’ll do the graphing and the same method for 28

i just need 23-26

[glass badger]

You use the same methods for questions 8 and 7

Complete the square

[topaz sinew]

@spring ivy Has your question been resolved?

[topaz sinew]

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[deep creek]

how does the highlighted step work, like why does it equal -1

[glass badger]

Recall the range of sinx

?<sinx<?

[deep creek]

0 -> 1?

or -1

[glass badger]

No

-1 and 1

[deep creek]

yes

[glass badger]

Now, what is the range of sin(2x+30)?

[deep creek]

im not sure ;-;

0 -> 360?

[glass badger]

There is only a change in phase and frequency, not amplitude

Look at it in relation to y

[deep creek]

is that the period that im trying to find

[glass badger]

What happens to the sinx graph function if you write sin2x, sin4x, sint 6x, etc etc?

The angles complete full 360 rotations faster

And so the frequency gets faster

And the graph gets squished sideways

But does this change the peaks?

No, it only reaches the peaks faster

That's why the range stays the same

[deep creek]

ohh

so theres just more values in the same range or

[glass badger]

No more values in the range, it still is -1 and 1

[deep creek]

okok

[glass badger]

Now which extreme would make y bigger, -1 or 1?

Keep not that you're multiplying by -4

-4 times -1 or -4 times 1?

[deep creek]

-1

[glass badger]

Yep

So sin(2x-30)=-1

You can solve this now

[deep creek]

okok

so i find the reference angle and do the unit circle stuff

[glass badger]

No need

At what angle does sin equal -1?

[deep creek]

-90?

or 270

[glass badger]

Yep

So sin(2x+30)=sin(270)

[deep creek]

and the thing given is 0<x<360

do i have to change that

because the answer has 2 values

so like 30<2x+30<2(360)+30

[glass badger]

No you don't need to do that

I'll explain what to do with it shortly, first, solve for x

[deep creek]

ok so 2x+30 = 270 right

[glass badger]

Not quite, you're missing some things

[deep creek]

then x = 120

[glass badger]

Note that sine is a periodic function

2x+30=270+360k

where k is a natural number

[deep creek]

but isnt it easier if u just do 2x+30 = 270

and then solve for x like that

[glass badger]

Sure, it's easier. But it won't give you all the answers

Sine repeats ever 360 degrees, but note that when you divide by 2 to solve for x, you're also dividing 360

That gives room to a second solution

X=120+180k

Now you try out possible values of k

This is where 0<x<360 comes in

[deep creek]

ohh okok

[glass badger]

If k= 0, what does x equal?

[deep creek]

120?

[glass badger]

Yes

And if k=1?

[deep creek]

300

[glass badger]

Yep

[deep creek]

OH okok i understand this now thank u

[glass badger]

If it was 0<x<180, then k=1 would have been a wrong solution

Np

[deep creek]

yep

alright thank u for ur help

.close

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[little cloak]

while doing completing square why the leading co-effiecent should be 1?

[hazy pumice]

It comes from what completing the square is

(Worst answer ever)

The visual proof and the algebraic derivation of quadratic formula

[inner oracle]

not necessary

[hazy pumice]

the leading coefficient is divided out

[little cloak]

are u sure?

[little cloak]

I have seen the geometric proof and how to do the process but just curious

[inner oracle]

yes

[hazy pumice]

You can always factor out the leading coefficient and then you have different b and c term

[little cloak]

so, it's not necessary to make leading co-efficent 1?

[hazy pumice]

It is

It won't work otherwise

[inner oracle]

[little cloak]

okay, it's all because of that geometric proof we have to isolate the co-effiecent

?

[hazy pumice]

@inner oracle you want to get in the weeds with this one? I am a little distracted rn tbh

[keen venture]

We're trying to take advantage of this identity:
x² + 2kx + k² = (x + k)²

[hazy pumice]

Perfect square trinomial ^

[keen venture]

I'm sure there's a similar identity that includes a leading coefficient, but why make it more complicated?

[inner oracle]

kaynex probably

[keen venture]

We can just factor that out, and use our simple method

[little cloak]

I mean when we divide something we get fraction which is more complicated?

[hazy pumice]

Watch this

ax^2 + bx + c

Factor out a

a(x^2 + (b/a)x + (c/a))

[keen venture]

You would still get fractions, but they would appear mid-process

[little cloak]

oh

[hazy pumice]

But now you still have a polynomial with 3 terms

a(x^2 + bx + c)

And we are back to the same problem anyway

In the general case

[little cloak]

For the last time I Am asking it's necessaly i mean purely necessary to have co-effecient 1 while doing perfect square process ?

[keen venture]

Well, that's what the perfect square process is, haha

So, uh, yes

[little cloak]

okay

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[coral fog]

Hello I'm not really good in factorization can someone help m

I get lost

the first line is where i have struggle

[summer wolf]

is the second thing muliplied by 2x-4

on the numerator i mean

[coral fog]

sorry can you repeat

[summer wolf]

does the numerator look like this

[coral fog]

yes

the 2

should be outside but she added parenthesis

[coral fog]

well, im not really good in factorisationm

[summer wolf]

that's because the factors are long

just imagine them as just an "x"

[coral fog]

but how did you do it

to get to -1+4(x-2)^2

[summer wolf]

ok let me show you

[coral fog]

can't I like expand?

[summer wolf]

at the beginning??

[coral fog]

yea right away

[summer wolf]

why

[coral fog]

idk factorisation is hard

I always have trouble

is there a trick I can always do to always get it right

idk from where to start

[summer wolf]

wait

[coral fog]

take your time

[summer wolf]

i forgot to add x^2 - 4x to this one so just ignore this

[summer wolf]

this is right anyhow

do this one make it more understandable

[coral fog]

yes

just one question

this is way better

you gave the minus of -8 to the a

in the third line?

and combined (x-2) together to get (x-2)^2

[summer wolf]

yes

[coral fog]

what about the 8

8(x-2)

second line

[summer wolf]

the 8 is a common factor

so i took it out

there are two (x-2) so that makes (x-2)^2

and then two (2) so 2x2=4

[coral fog]

yes

[summer wolf]

do you understand it now?

[coral fog]

it's only the 8

where did it dissapear

[summer wolf]

ok let me expand my working a bit more

[coral fog]

thank you

[summer wolf]

capeesh?

[coral fog]

in the third line we removed one "a" and took common factor another one?

right?

[summer wolf]

could you please rephrase that

ohhh

wait

yes

yes

[summer wolf]

yes

[coral fog]

so we factorized one

and took one as common factor

[summer wolf]

well you can take both 8 and a out at the same time

but i was just showing you in more detail

[coral fog]

I think I got it

thank you

[summer wolf]

np

you get better at this when you practise more questions

[coral fog]

but expanding it won't get me the same answer?

[summer wolf]

it will

if you've done it correctly

[coral fog]

but I have to factorize in the end aswell

to get exactly the same answer

[summer wolf]

oh you mean at the beginning

[coral fog]

she wants exactly the same answer

yes

[summer wolf]

i would not suggest expanding at the beginning

such a waste of time

[coral fog]

Okay

ty

if I have an 8 outside as common factor and one inside

I remove one of them/

?

[summer wolf]

well if you take out the 8 then you remove the inside one

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[random nexus]

hi

[topaz sinew]

@random nexus Has your question been resolved?

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[nova tinsel]

I have a 7th grade geometry question so it’s probably easy for you guys but anyway. Here is my problem, and I need help how to work it. Our notes are terrible.

[nova tinsel]

I Will send picture now

I have to use a1 over a2 = s1 over s2

Im about to go to school I need help 💀

[pallid phoenix]

if the two traingles are simillar the proportion between the same sides are equal

for exmple, 6.25/5 = x/8

in ur picture

if only the two triangles are simillar

[nova tinsel]

Ok thanks so much

[nova tinsel]

Also though

Which one do I square because we said we had to square it

[pallid phoenix]

sqaure what

[nova tinsel]

Normally it would be a1/a2 =( s1/s2 ) squared

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[rain radish]

What is the question

[viscid saffron]

Find the center and radius of the circle equation

I already did it, here is the evidence:

The center is already good

but the center i'm not sure

Cause I need to simplify

@rain radish are you there?

[compact sigil]

Do you know what completing the square is

[viscid saffron]

yes

[compact sigil]

Is it the radius you still need?

[viscid saffron]

yes

the center is already good

[compact sigil]

You square root the value on the right hand side of the equation

The 89/4

[viscid saffron]

yeah?

[compact sigil]

You square root it to get the radius

[viscid saffron]

cause they told me it was 69/4

[compact sigil]

Who told you it was 69/4

[viscid saffron]

my teacher

but not sure about it

Cause when I do 1 + 9 + 49/4 it gives me 89/4

[compact sigil]

That’s what I got

[viscid saffron]

and not 69/4

[viscid saffron]

not to simplify the radius

is it possible?

[compact sigil]

The simplification would be root 89 over 2

[viscid saffron]

yeah like that

[compact sigil]

That’s it in it’s simplest terms

[viscid saffron]

ok I got it

Thank you so much @compact sigil

btw can I add you as a friend?

[compact sigil]

Yeah sure, and no problem! :))

[viscid saffron]

ALRIGHT!

Oh also if you could help me solve also these problems which are related to the same topic:

I will do it also right now but I will show you if my procedure it's correct 🙂

[compact sigil]

Yeah no problem

[topaz sinew]

@viscid saffron Has your question been resolved?

[viscid saffron]

@compact sigil I already did it

the second point

the radius I don't know what it is

@compact sigil

[compact sigil]

That’s what I got

[viscid saffron]

but wait

is 1/3 or 1/6?

cause the center I got (1/3,-1/4)

@compact sigil

[compact sigil]

If you divide 1/3 by two you get 1/6 you forgot to do that when completing the square

[viscid saffron]

wdym?

if (1/3)^2 is 1/9 from where 1/6 does it come from?

[compact sigil]

When you complete the square you remove the x from the x term and divide by 2

[viscid saffron]

wait let me check

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[deep heron]

idk how this works but hope im doing this right: probably a really basic question but whats the chances of throwing a pair with 5 dice? (exclusively any pair and the other 3 dice being distinct from one another / unique)

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[distant lintel]

Can someone help me understand the work for this problem?

[distant lintel]

We got that the series is 5/4, which is greater then 1 therefore it diverges

but I'm a little confused on the algebra and simplification, especially for the factorials

[alpine mist]

like which part?

[distant lintel]

Do you see the 2nd two red circled parts?

The one with (2n)!/(2n+2)!

[loud ingot]

What is Thomas Sowell doing here?

[distant lintel]

I do not understand how the denominator turns into (2n+2)(2n+1)2n!...perhaps I still do not understand how factorials work..

[alpine mist]

k! = k * (k-1)!
your 'k' here is 2n+2

[loud ingot]

Eaz as fuck bro

[alpine mist]

so (2n+2)! = (2n+2) * (2n+1)!
do it again to (2n+1)!

[loud ingot]

5! = 5 * 4 * 3! For example

I thought Sowell was smarter

[distant lintel]

So (2n+1)! = (2n+1)*2n?

[distant lintel]

i just didn't pay attention to this in class, finals have had me drained.

[alpine mist]

well, (2n+1) * (2n)!

[distant lintel]

Right, do you mind if we go over a ratio test problem together?

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[alpine mist]

if you want to work one out and let me know when you're stuck, sure.

[topaz sinew]

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[keen solstice]

Help

[hardy wing]

Do you know what a Pythagorean triple is?

[keen solstice]

I think i do

im just confused because this problem the numbers have square roots on them and im not sure what to do

i figured it out

4=7

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[frosty heart]

How would I go about 13. a)? I know it has to do something to do with spanning sets in just not sure how to go about it.

[topaz sinew]

@frosty heart Has your question been resolved?

[frosty heart]

I got as far as a(-1,2,3) + b(4,1,-2) = (-14,-1,16) and then -a+4b=-14 , 2a+b=-1 , 3a-2b=16

[topaz sinew]

@frosty heart Has your question been resolved?

[frosty heart]

<@&286206848099549185>

[frosty garden]

Let’s recall some equivalent way of saying three vectors does not lie on same plane

1. They are linearly independent.
2. The volume of the parallelpiped formed by these three vectors is not 0.

Maybe we can start from (2), can you recall how to compute such volume? And how does it lead to your desired conclusion?

[frosty heart]

Wait do I just have to solve for the scalar multiples and have my my LS=RS check to fail?

[frosty heart]

Ah I got it

.clos

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[lapis iron]

Hello

[lapis iron]

Can anyone help me find the variance for (b)

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[icy lagoon]

How do i do this?

[topaz sinew]

@icy lagoon Has your question been resolved?

[icy lagoon]

<@&286206848099549185>

[icy lagoon]

<@&286206848099549185>

[icy lagoon]

<@&286206848099549185>

[vernal vale]

wish i could figure out how to approach this

i was trying set f'=0

[icy lagoon]

yeah>

?

[vernal vale]

sorry trying desperately not to make stupid algebra mistakes

solving f'=0 you get a form for x

then uhh

f of that x should be 4/3 or 2/3

i guess and, really

[icy lagoon]

yea

man idk

i can get an expression completely in terms of a

and thats all fine but idk how to turn it into a range of possible values

[vernal vale]

theres only one solution

[icy lagoon]

or how to set just the bounds of the function as the global mac

max and min

oh ok

yeah will ive got

[vernal vale]

somewhere around -sqrt(2)

[icy lagoon]

ive got this

[vernal vale]

,rotate -90

[icy lagoon]

ignore the quadratic bit its the same answer

[vernal vale]

simpler to work with $x=\frac{1}{\sqrt{-3a}}$

imo

[thorny flame]

jan Niku

[vernal vale]

does this get you to $\mp \frac{1}{3\sqrt{-3a}} \pm \frac{1}{\sqrt{-3a}} + 1$

[thorny flame]

jan Niku

[icy lagoon]

what is happening

so

how do you get $x=\frac{1}{\sqrt{-3a}}$

[thorny flame]

Wilsonkhann

[vernal vale]

inspection

[icy lagoon]

oh you square root both

ok

ok cool

now what do i do with that

[vernal vale]

the same thing you did with your x i think

im not sure its any nicer

plug it into f

try to simplify it

[icy lagoon]

yeah hangon

i might try the bounds first i think it works

[vernal vale]

i have to fall asleep or im gonna die

if you do manage to solve this congrats, seems really annoying

[icy lagoon]

lmao allg

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[lone nest]

(Pssst. Instead of the internal-local-extrema solution, start by looking for an a that makes f(-1)=4/3 and f(1)=2/3, and then check whether the endpoints of that are actually global max and min on [-1,1] ...).

(Then look for an a that makes f(-1)=2/3 and f(1)=4/3)

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[empty sail]

You need a new channel

This will close

[empty sail]

Because of this

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[neon iron]

Who can help me with my ixl

[neon iron]

Ive been on it for a total of 6 hours and haven’t passed yet

[simple orchid]

What's your question?

[neon iron]

One sec

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@neon iron Has your question been resolved?

[violet holly]

[neon iron]

Its not uploading

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[neon iron]

I got another one

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[main ibex]

✓(16-x^2)+✓(9-x^2)=5

[main ibex]

How to solve this

We can square both sides

[thorny flame]

AustinU

[rigid ivy]

Let u²=9-x², then your life is a tad easier

[main ibex]

I calculated and at last I got
2x^4-27x^2=|144|

[rigid ivy]

!show

[topaz sinew]

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

[main ibex]

By squaring both sides

[prisma mesa]

you messed it up somewhere

Squaring both sides isnt bad idea, as long as you know that it might generate extraneous solutions. It will make it really easy because 16+9 will cancel with 5^2. You just messed up the algebra

[main ibex]

Alr Im redoing it

[ionic oar]

Maybe

You should plot the graphs for $f(x) = \sqrt{16 - x^2}$ and $g(x) = 5 - \sqrt{9 - x^2}$

[thorny flame]

NEONPerseus

[ionic oar]

They're both semi circles and pretty easy to plot

Wait the second one isn't one is it

,w plot y = 5 - \sqrt{9 - x^2}

[thorny flame]

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[main ibex]

I solved it

[ionic oar]

Good on you

[main ibex]

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[neon iron]

Need help with my gcse maths revision, got 6 days, if anyone could tutor me a bit too I would appreciate it very much

Stuck on this question

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@neon iron Has your question been resolved?

[haughty wren]

well @neon iron, your goal is to isolate the variable. You have to think in an operation kind of way. Let's try it on 4y + 5 > 12. First thing, look at additions and subtractions. There's an addition on the y side, so since 5 is being added, you need to subtract it from both sides giving you: 4y + 5 - 5 > 12 -5. Now we now 5-5 = 0, and 12 - 5 = 7 giving: 4y > 7. Next, we see that there is a multiplication next to the y. Since the 4 is multiplying, let's divide it on both sides giving: 4y / 4 > 7 / 4. 4/4 = 1, meaning: y > 7/4. You could enter 7/4 in your calculator, but it would give an approximate answer, so I'd suggest leaving it at that. It's not all that complex when you think about doing the opposite of what's happening on the variable's side.

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[empty sail]

Don't say that you need help. Just post your question

Don't ping me

[loud oasis]

there are 3 trigonometric ratios to consider: sin, cos, or tan

compared to the angle we're given, which two sides do we have (out of opposite, adjacent, hypotenuse)

which one is which

[empty sail]

The trig function you use is based on the perspective of where the angle is

[loud oasis]

adjacent means it's touching the angle

[empty sail]

So is 18 opp or adj from the angle?

Yes

So what trig function should you use?

[loud oasis]

we've established that we have opposite and hypotenuse. So what ratio involves those two sides?

so let's write out the sin ratio, plugging in the values we know

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[uncut agate]

If a, b and c are positive constants show that all solutions of the DE ay''+by'+cy = 0, tend to 0 when x --> infinity.

I don't how to do this, my equation is ar^2 + br + c = 0, then I have 3 cases, when b^2 - 4ac > 0, my solution is y(x) = c*e^r1x + de^r2x, if I calculate the limit x ---> infinity, it's equal to infinity no 0, and the other cases are similar, then what I have to do?

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[blazing crescent]

how do i work this out?

[abstract wadi]

Chain rule.

[blazing crescent]

(cosec(x))^3?

the differenciate?

yh i see it

thanks

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[dense nova]

we finished derivatives a while back and then started integration, my teacher repeatedly says unlike derivatives integration is much more limited and differentiation is possible in a lot more cases unlike integration
When we studied derivates before that we studied differentiability so I could actually tell if a function is differentiable at some point in its domain, how can i say the same for integrals?

[dense nova]

basically when/how can I conclude a function is non integrable?

[halcyon slate]

There are different notions of integrability, for the most basic ie Riemann integration. A function is Riemann integrable if the measure of its set of discontinuities has measure zero

For a simpler case, if the function is discontinuous at finitely many points then it is Riemann integrable

[dense nova]

Something like sin(x)/x, that is?

[halcyon slate]

Wdym?

[dense nova]

sin(x)/x even though is discontinuous at x = 0, it is integrable?

In an interval that includes 0

,w int sin(x)/x from -1 to 1

[thorny flame]

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[dense nova]

...

what's that

[halcyon slate]

sinx/x is also called the sinc function

And since it is an even function the integral can be simplified

[halcyon slate]

Yes although it's not a proper integral if the limit involves infinity

[dense nova]

the limit x to 0 is actually 1 tho?

also what exactly do you mean riemann integrable

[halcyon slate]

You know how integral is taught as taking rectangles and adding their area. Then in the limiting sense the sum of the area approaches the integral. This notion of integration is called Riemann integration

Then there's a more advanced notion of integration which uses the concept of measures to define integration. It is called Lebesgue integration.

[dense nova]

didn't Newton/leibniz found integration, is their notion not that?

[halcyon slate]

What they did was in an intuitive sense. The proper theory of integration took much longer to form and is different than what they had in mind. In fact the concepts of limit that we use to define integration cause after Integration.

It gave birth to whole new areas of maths called real analysis and measure theory

[alpine mist]

usually when teachers say this they mean the methods we have for finding an antiderivative are more limited than derivatives. We can find the formula for the derivative in a lot more cases, but it's very easy to find something that you can't find a simple antiderivative for, or where basic methods break down.

[dense nova]

yes, we can always pretty much use the first principles for derivatives, is there no such thing for integration?

i guess not

[alpine mist]

like, the integral for e^(x^2) requires some more advanced techniques than 'power rule' or 'by parts' methods.

[dense nova]

can't you expand it and then integrate it?

and then find some pattern

or

...

[alpine mist]

,w integral of e^(x^2)

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[dense nova]

that's not so good

i get what you all mean

one last question

when would someone typically see an integral like e^(x^2)

And this result, 1/2sqrt(pi)erfi(x)

(whatever the heck that means)

[halcyon slate]

You'd find some interesting integrals while doing advanced physics

[dense nova]

not maths?

[halcyon slate]

Not much unless you're solving differential equations

[dense nova]

i did find some integral in physics last session that apparently had something to do with gamma function and so we had to skip it

[dense nova]

i see

alright!

thank you both!

.close

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[gleaming thunder]

see elliptic integrals as well as the error function

[dense nova]

.reopen

[topaz sinew]

✅

[dense nova]

any source you'd recommend

[gleaming thunder]

elliptic integrals appear when studying pendulums for example, while erf describes heat flow in simple cases

[gleaming thunder]

not really anything no. These are things we talked about in class, and that very recently came up in a french exam paper

[dense nova]

which exam paper?

[gleaming thunder]

Mines-Ponts 2023 MP physique 1

[dense nova]

nice

[gleaming thunder]

the paper decided to take classic problems for which you usually approximate, and use some more heavy machinery to do them without approximating

using the lambert W function, elliptic integrals and the error function in 3 separate studies

free falling body with friction, period of a pendulum, heat flow

[dense nova]

also talking about simple problems, i remember a physics problem which deals with electric fields, would you mind if I shared it here/expressed your views on it

of course I'll take this elsewhere as well later today, as this isn't a physics server

[gleaming thunder]

I'm not much of a physics guy but feel free to go ahead

[dense nova]

it has something to do with electric field at a point which is not at the centre, but very near it, like it approaches the centre you might say, the centre is of a uniformly charged ring

[gleaming thunder]

cant you send an image ?

[dense nova]

this question was given to me by my teacher once I asked him to give an expression for electric field by the same ring at a point near the ring
So basically there's no original problem, let me explain
we were studying electric field, and so we studied electric field due to a ring with uniform charge distribution
Then he told us it is supposed to be zero at centre and we wrote an expression for electric field at its axis
Later I asked about the electric field at some point on ring, if that makes sense
And he said it isn't possible to do that but it is definitely possible to find gje electric field at a point just near the centre
I claimed its zero but he said its only zero at the exact centre, nowhere else and he kind of makes sense, so that is the question he has given me for the time being
I'm sorry I don't have any image

[gleaming thunder]

what do you mean by ring more precisely ?

a flat one with a width, a cylinder with a thickness, just a circle ?

[dense nova]

a Circle

simply

[gleaming thunder]

with lineic charge density ?

[dense nova]

yes

Uniform linear charge density

[gleaming thunder]

a charge generates potentiel q/r
the potential at a given point would be the integral of q/r over the circle
then you need to find r(theta)
let the circle have radius R, the point be a distance d from the center
r² = x²+y²
(x-d)²+y²=R²
r² = R² - d² + 2dx
x = d + R cos theta

,w integrate 1/(k + cos theta) from 0 to 2pi

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[gleaming thunder]

not exactly nice

but usable

[dense nova]

how's (x-d)^2+y^2 = r^2

[gleaming thunder]

I center the point at the origin, and offset the circle

[dense nova]

isn't that an approximation already?

[gleaming thunder]

no

,w integrate (q / sqrt(R²-d² + 2d(d + R cos theta)) dtheta) from 0 to 2pi

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[gleaming thunder]

,w integrate 1 / sqrt(k + cos(theta)) wrt theta from 0 to 2pi

[dense nova]

the point was near the centre but you wrote the equation of the circle considering the centre was located at that point

[gleaming thunder]

goddamit

that's bad

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[gleaming thunder]

let the point be at (0, 0)
the circle is centered at (d, 0)

[dense nova]

i see, okay. that's that.

[gleaming thunder]

and I wanted to differentiate that wrt d afterwards

that's not gonna work

[dense nova]

but it is doable?

[gleaming thunder]

algebraically ?

[dense nova]

any possible way

i don't have a preference for algebra or anything

[gleaming thunder]

computers can always find things to arbitrary precision

in practice

but if you're going this way, might as well compute the field directly

wait

that removes the sqrt doesnt it

[gleaming thunder]

the field is q/r² er
er = cos theta ex + sin theta ey
then you integrate
q/(R²-d² + 2d(d+R cos theta)) cos theta for the x component

,w integrate cos theta / (k + cos theta) wrt theta

not pleasant

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[gleaming thunder]

,w integrate sin theta / (k + cos theta) wrt theta

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[gleaming thunder]

so technically you can find the field. But that's not exactly elegant

especially plugging in the value of k and finding the outside constant

makes it even uglier

[dense nova]

i see, I'll try it in this way and see where that leads me

of course I'll also be using software for most of the integration

[gleaming thunder]

to pain and suffering according to all available evidence

[dense nova]

well this question is merely a fun problem so i can send my teacher anything of the sort its not like im doing the hard work anyway haha

[gleaming thunder]

also don't forget the 1/4pi e0

but yeah very ugly

[dense nova]

ty for all your efforts

.close

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[undone flicker]

Any idea to solve it with series?

[drifting swift]

do you even need series? direct substitution works just fine here

sin( (y-a)/2) approaches sin(-a/2) and tan( (πy)/(2a) ) approaches tan(0) = 0

[topaz sinew]

@undone flicker Has your question been resolved?

[undone flicker]

Tan 0 how?

Can you please explain how it approaches? Did you put the value of y directly?

[sweet shard]

,calc tan(0)

[thorny flame]

Result:

0

[sweet shard]

this is a standard fact from trigonometry

[undone flicker]

What do you mean say clearly?

I am not asking value of tan 0
My question is how tan 0 comes here if we put y = a it becomes tan pi/2 which is infinity

[sweet shard]

you're doing this limit problem right?

[undone flicker]

Ohh wait

It's 0 i thought it's (a)

So what next step

Here we have got sin(-a/2)× tan 0 which is zero so all value will be 0

Zero is not the answer

[sweet shard]

answer to what

[undone flicker]

To this

[undone flicker]

What will be the next step

[undone flicker]

??@sweet shard

[sweet shard]

why do you think 0 is not the answer

[undone flicker]

They says
Sin (-a/2) . Tan0

It becomes 0 due to tan0 no??

[sweet shard]

don't understand what you're saying

Here we have got sin(-a/2)× tan 0 which is zero so all value will be 0
this is correct

[undone flicker]

Sin(-a/2) × tan 0 = sin(-a/2) × 0 =0

[sweet shard]

so why do you say zero is not the answer?

[undone flicker]

Because zero is not the correct answer

In answer sheet

Understand???

[sweet shard]

show the answer sheet

screenshot / picture is best

[sweet shard]

.

[undone flicker]

I guess the limit confusing

[sweet shard]

these two are different problems

[sweet shard]

this is not "answer sheet"

[undone flicker]

Yeah but look at the question/solution

They added a different answer sheet to the question

[sweet shard]

it's a different question

[sweet shard]

read.

[undone flicker]

Yes got it thats the confusion

Even I didn't notice it while reading the solution

So tell me now how to solve it when limit tends to a

I meant what approach?

[sweet shard]

they explain the answer to you

l'hopital is the main tool

[undone flicker]

No i don't want to solve it through this as I have clearly mentioned here

I want to solve it with any other method if it has an expansion of a series of something

[sweet shard]

that was for a different problem

[sweet shard]

then use taylor series up to order 2 or 3 for both sin and tan

[undone flicker]

Yes. It was

I tried it already but it becomes 0

[sweet shard]

Show

[sweet shard]

,rotate

[sweet shard]

Yea I can't follow what you did

,w Taylor tan(z)

[undone flicker]

Ohh wait a minute then. It's messy

[thorny flame]

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Requested by riemann#2529

[undone flicker]

Yes. I followed this

Sinx and tanx i put value of x too

Then I multiply it but it becomes zero

[sweet shard]

You just did algebra wrong probably

[undone flicker]

I see, let me try it again clearly then show you

It was messy so maybe my mistake

@sweet shard

Is this correct? Please check

[topaz sinew]

@undone flicker Has your question been resolved?

[topaz sinew]

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[merry mason]

If n(A)=4, n(B)=5 and n(A∩B)=3, then n((AxB)∩(BxA)) is?

[merry mason]

Is answer 8 or 9?

[neon venture]

n is cardinality ?

[merry mason]

Cardinal number, yes

no of elements in set A is what my teacher call cardinal number of set A, represented by n(A)

[narrow flint]

What is the title of the lesson?

[merry mason]

Relation and functions lecture no 8

Sir was doing test revision

And it was a question asked in that revision

So no specific title for lesson

Also it is not this specific question

[narrow flint]

I think I've solved a question like this before

I will search for u in my notebook

[merry mason]

The question asked was if n(A)=3, n(B)=4 and n(A∩B)=2 then n((AxB)∩(BxA))=

I simply edited the original question

[merry mason]

No worries

I got answer through lengthy example building

Answer is 9

[narrow flint]

Did u find a detailed answer?

[merry mason]

Yes I guess, it was through example building but it's one lengthy method

I have a new question though.

If I have 3 elements in a set and I want to make every possible pairs out of these 3 elements, how would I do that?

[narrow flint]

I also found the solution in my notebook, but I did not find how to explain it to you because my lessons are not in English, sorry

[merry mason]

It's okay

I'll ask new question from fresh channel

.close

[topaz sinew]

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[merry mason]

If you are picking every possible pairs out of 3 elements then 3p2 which is 6 is not the answer because this doesn't make (a,a) , (b,b), and (c,c) pairs btw

[merry mason]

@safe burrow I think I needed to tell you this

[safe burrow]

so it's pair drawing w/o replacement?

[merry mason]

Yes

I kind of missed it that's why I said 9 because from another method that was the answer

[safe burrow]

for a set with n elements, that's just n(n-1)

[merry mason]

Ohh

which is still 6 though, because n=3

[safe burrow]

since you have n choices for the first element, and (bc you cant pick that again) n-1 choices for the next

and last

[merry mason]

Shouldnt be n^2 though

[safe burrow]

yeah. if your choices are a, b, c, you get
ab
ac
ba
bc
ca
cb

[merry mason]

And aa, bb, and cc

Don't we consider them as well?

Permutation function completely missed them

[safe burrow]

i thought you couldn't pick the same element twice?

was that not the whole point?

[merry mason]

If we are making Cartesian product we can do that

[safe burrow]

if we can have aa, why did you say yes here?

[merry mason]

It's also my problem, I forgot to mention you that we can pick same elements as well 😔

Ohh

I couldn't comprehend that well then which is why I said yes

Now I understand

Also thanks for your help, I have fixed this now

.close

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[neon iron]

girll im royalty

[neon iron]

bow down

😭

but srsly i need to say thx to everyone who helped my pass my final

huges

[knotty ledge]

does her majesty have a maths question

[neon iron]

https://tenor.com/view/hello-gif-26209356

[radiant tapir]

kk I'mma close this

.close

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[topaz sinew]

@digital sleet Has your question been resolved?

[tropic fern]

@digital sleet yo

what have you tried on this one

[digital sleet]

lmao sup again

[tropic fern]

sup xD

[digital sleet]

soo yea im not quite sure even how to approach this one

is the trapezoid cyclic?

[tropic fern]

you will have to answer that yourself, since it's not given then assume it's not at first, you have two options it's either touching on the bases or on the sides

[digital sleet]

im also not too sure about the properties of a circumcircle

[tropic fern]

hmm thinking about it, would be dumb to have a circumcircle who doesn't touch all vertices

it is safe to say circle touches every side of the trapezoid @digital sleet

[digital sleet]

this is just a guess but would a line connecting two opposite endpoints be the diameter?

[tropic fern]

it would yes

[digital sleet]

cause ik thats the case for a right triangle but idk bout an iscoceles trapezoid

alr

[digital sleet]

wait thats an incircle

[tropic fern]

you know the sides, you know the bases, find x, and solve for height

[digital sleet]

the height is 4sqrt2

[tropic fern]

oh well

i apologize i was thinking it was inscribed all along

you need to find it's diagonal

always draw a picture

[elfin sparrow]

The diagonal isn't the diameter of the circle, is it?

[digital sleet]

idk ig thats what we're assuming

[elfin sparrow]

If it is, then you've got a rectangle, not a trapezoid

[tropic fern]

nop it's not 💀

[digital sleet]

oh then what do we do lmao

[tropic fern]

so middle of the trapezoid base to it's touching side

is a radius?

nop

not it aswell

lemme draw some more

[digital sleet]

circumcenter to a point is

but idk an properties bout that lmao and google isnt giving me any

[elfin sparrow]

I've tried a few things but I'm not really sure

I tried assuming the base of 10 was the diameter but that's definitely not true

[tropic fern]

goddamn should have learned more about trapezoids

@digital sleet what are these questions from

[elfin sparrow]

When you draw the diagonal, you make two triangles, and the circumcircle of the trapezoid is also the circumcircle of both triangles

which means the point where any of their perpendicular bisectors intersect is the center of the circle

I have to go in a minute but that seems like a path

[digital sleet]

mathcon

[topaz sinew]

@digital sleet Has your question been resolved?

[digital sleet]

.close

[topaz sinew]

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Closed by @digital sleet

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

[undone flicker]

.reopen

[undone flicker]

Hello again. It's closed by both. You didn't respond

[topaz sinew]

Channel closed

Closed due to the original message being deleted

[undone flicker]

@sweet shard

[undone flicker]

@drifting swift

[empty sail]

Don't ping specific people

[undone flicker]

Here is another question but the limit changed only. This is what I tried

@empty sail we were trying to solve

Above but closed the question by bot

[undone flicker]

Here

This is what I tried

[empty sail]

That was hours ago

[undone flicker]

Yes. So @empty sail

I just woke up

[empty sail]

So, don't ping specific people