help-33

[ebon depot]

$x-1 = Ax^2 + Ax +Bx + B -\frac23 x^2 -\frac43$

[elfin berry]

Merineth

[fervent rampart]

now we'll group all the x² terms, all of the x terms, and all of the constant terms

[ebon depot]

but

Can't i just put in x = 0

that way A is gone and we get B

B = 4/3 - 1

B = 1/3

[fervent rampart]

that works

[ebon depot]

$x-1 = Ax^2 + Ax + \frac13 x + \frac13 -\frac23 x^2 -\frac43$

[elfin berry]

Merineth

[fervent rampart]

the general method for this is:

1. expand:
   $x-1=Ax^2 + Ax + Bx + B + Cx^2 + 2C$
   group all like terms into a single coefficient:
   $0x^2+1x-1 = (A+C)x^2+(A+B)x+(B+2C)$

[elfin berry]

cloud

[fervent rampart]

by comparing coefficients, we find the following must be true:

1. x² coefficient must be 0:

$0 = A + C$

2. x coefficient must be 1:

$1 = A + B$

3. constants must add up to -1:

$-1=B+2C$

[elfin berry]

cloud

[ebon depot]

I think A = 2/3

[fervent rampart]

you can either solve this as a system of equations from the start (hard but doable) or you can use substitution to make it simpler (in this case we know everything except A)

[ebon depot]

Yeah!

I put in B and C

and then i saw that i could make x = 1

Giving me an easy equation

$\frac{\frac23x + \frac13 }{x^2+2} + \frac{-\frac23}{x+1}$

[elfin berry]

Merineth

[ebon depot]

That should now be in partial fractions

[fervent rampart]

we now get back to the integral: remembering to add 1 to the front from our earlier division, we can now integrate this

[ebon depot]

oh nearly forgot the +1

$\int 1 +\frac{\frac23x + \frac13 }{x^2+2} - \frac{\frac23}{x+1}$

[elfin berry]

Merineth

[ebon depot]

can't i make these 3 seperate integrals?

or rather 2 since 1 is just 0

[fervent rampart]

we can also split

$\frac{\frac23x + \frac13}{x^2+2} = \frac{\frac23x}{x^2+2} + \frac{\frac13}{x^2+2}$

[elfin berry]

cloud

[fervent rampart]

now we split into 4 different integrals and solve (mostly by substitution)

[fervent rampart]

i think you're confusing integrals and derivatives

[ebon depot]

Woops, yeah

I'll do those calculations on paper

gets a bit tricky with the bot

I got a question about F(x) tho

if i have

$f(x) = \frac{2x/3}{x^2+2}$

To find F(x)

do i just take 2/3

and find primitve of that

[elfin berry]

Merineth

[ebon depot]

or do i have to make

$(2x/3)(x^2+2)^{-1}$

[elfin berry]

Merineth

[ebon depot]

and then find primitive of these two ?

[fervent rampart]

this one is solvable by substituting u = the denominator

[ebon depot]

oooh

1/u

1/x = ln|x|

but

is 1 substitued with 2x/3 ?

[fervent rampart]

remember that when we're substituting, we have to divide by the derivative of u

[ebon depot]

$\int \frac {2x/3}{x^2+2} = \int (2x/3)ln|x^2+2|$

[elfin berry]

Merineth

[ebon depot]

So this is wrong?

[fervent rampart]

that is wrong

[ebon depot]

:(

I'm not too familiar with substitution. Why do we have to divide by the deriviate of u?

$\int \frac {2x/3}{x^2+2} = [(1/3)ln|x^2+2|]$

I'll just remember to do it

[elfin berry]

Merineth

[ebon depot]

So i get

$x + \frac{ln|x^2+2|}{3} + \frac{1/3ln|x^2+2|}{2x} - \frac{2ln|x+1|}{3}$

[elfin berry]

Merineth

[fervent rampart]

because substitution is the chain rule in reverse

[ebon depot]

I had a suspicion it had something to do with the chain rule :3

[ebon depot]

can i go even further than this?

[fervent rampart]

that's nearly everything, we still need to integrate (1/3)/(x²+2)

[ebon depot]

aaaaaaaaaaa i missed that one

One moment haha

$x + \frac{ln|x^2+2|}{3} + \frac{1/3ln|x^2+2|}{2x} - \frac{2ln|x+1|}{3}$

[elfin berry]

Merineth

[ebon depot]

That should be it

[fervent rampart]

unfortunately that's not a correct integral of (1/3)/(x²+2)

[ebon depot]

Is it not?

denominator becomes u

1/3ln|x^2 +2|

[fervent rampart]

when we have a u-sub we have to get everything in terms of u, with no leftover x

[ebon depot]

and divide by deriviate of x^2 which is 2x

$\frac{1/3}{x^2+2} = \frac{1/3}{u} = [(1/3ln|u|) / u']$

That would be my thought process

[elfin berry]

Merineth

[fervent rampart]

we would be left with (1/3)/u * 1/2x, with nothing to cancel the 2x with

when we have these types of integrals we can instead rewrite:

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

[elfin berry]

cloud

[fervent rampart]

that's our denominator. then you take out all the constants to the front and u-sub u = x/√2

[ebon depot]

I have no idea what that means

[fervent rampart]

we're trying to turn it into the form
1/(u²+1) because that integrates to arctan(u)

[ebon depot]

how do you know

that youy want to achieve arctan?

[fervent rampart]

if we have integral in the form 1/(x²+a) that's close enough to 1/(x²+1)

[ebon depot]

but arctan is

so you want to have a = 1 since 1^1 is 1

[fervent rampart]

that's a formula that includes the u-sub as a hidden step

you can just apply that formula directly

[ebon depot]

That is the only formula that provides arctan

[fervent rampart]

the derivative of arctan(x) is 1/(x²+1). by the fundamental theorem of calculus, the integral of 1/(x²+1) is arctan(x). the formula above is using that + the same u-sub i showed above to make a more useful/general formula

[ebon depot]

$[\frac{1/3}{2^2}tan^{-1}(\frac{x}{2})]$

[elfin berry]

Merineth

[fervent rampart]

note that in our integral a² = 2

so a must be √2

[ebon depot]

But it says a should be squared?

or nvm i guess i get it

$x + \frac{ln|x^2+2|}{3} + \frac{tan^{-1}(x/\sqrt{2})}{3\sqrt{2}} - \frac{2ln|x+1|}{3}$

[elfin berry]

Merineth

[ebon depot]

there we go..

[fervent rampart]

+C

[ebon depot]

I need to get some rest now before i die

this has to be the hardest one i've ever done

painfully hard

[fervent rampart]

none of the individual steps are especially hard (once you practice them) but partial fractions problems can get really long

[ebon depot]

I’ll just have to practice more tomorrow I think

Def need a break now

Thanks for the help ;-;

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[pearl zealot]

Yo

[pearl zealot]

I need help with prooving the complanarity of vectors

My vectors are A(1:0:1) , B(0:1:0) , C(2:1:2)

Now i dont know how i am supposed to handle the 0

like its 0r

so thats zero

but what then

Im trying to solve it with the gaus algorythm

So its
I r + 0 +2t
II 0 + s + t
III r + 0 + 2t

[marsh citrus]

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[hollow moth]

Hi, trying to solve this probability q
Suppose that two players A and B take turns rolling a pair of balanced dice and
that the winner is the first to roll at 7. If A rolls first what is the probability A will win?

I can arrive at (5/6)^(k-1) * (1/6) which is the probability where k is the number of roles but unsure of how to proceed from here.

[marsh citrus]

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[vagrant gull]

,, 2^{2x+1}-17\cdot 2^x=-8

[elfin berry]

Akira (fumo)

[vagrant gull]

I've began but stuck midway

ill show work

,, (2^x)^2 \cdot 2 - 17 \cdot 2^x = -8

let $u = 2^x$

[elfin berry]

Akira (fumo)

Akira (fumo)

[vagrant gull]

now it's $u^2 \cdot 2 -17u =-8$

[elfin berry]

Akira (fumo)

[vagrant gull]

then what

[pulsar sluice]

solve for u?

[vagrant gull]

yeah but im getting stuck

how they got u=1/2 and u=8?

[pulsar sluice]

2u^2-17u+8 = 0

[vagrant gull]

so $(2u-1)(u-8)$?

[elfin berry]

Akira (fumo)

[pulsar sluice]

yep

[vagrant gull]

oh alright

thanks mate

[pulsar sluice]

np

[vagrant gull]

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[late geode]

factor out log_5(x)

[marsh citrus]

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✅

[hasty ruin]

that would be incorrect

[late geode]

would there be no solution for when i say that the first factor = 0
how are you reaching that conclusion

5 - log_5(x)^4 = 0
log_5(x) = 5
log_5(x) = 5^(1/4)
you missed the power in that second line
just one more step after this

yep

where's 11 coming from

use approx instead of =

wdym by cancel the exponent

yes

use something like 4th-root or exponents instead of 4 next time

[marsh citrus]

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[sterile drum]

h=\frac{\left(100\right)^{2}\left(\sin30\right)\left(\sin30\right)}{2\cdot9.81}
h=497.556 (should be 127.4)

Why am i not getting the same result (using desmos)? 😅

[cobalt sentinel]

Maybe degrees vs radians

[sterile drum]

Edit, yeah its radians, thanks! :D But man thats messy written out by the person, this formula doesnt make any sence without adding Degree*0.0174533.

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[signal apex]

Is the answer B?

[signal apex]

I would like to know if the answer is B

[runic temple]

sounds good to me

[signal apex]

thanks

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[native frigate]

,w 10 = 15+22x-5x^2

[elfin berry]

Results provided by WolframAlpha

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

Requested by tuzis

[marsh citrus]

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[marsh citrus]

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[cyan loom]

hello i need help with excel math

[cyan loom]

.close

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[quartz gull]

Alina wants to make keepsake boxes for her two best friends. She doesn't have a lot of money, so she wants to make each box described so that it holds as much as possible with a limited amount of material.
For Jen, Alina wants to make a box with a square base whose sides and base are made of wood and whose top is made of metal. The wood she wants to use costs 5 cents per square inch, while the material for the metal top costs 12 cents per square inch. What is the largest possible box (in terms of volume measured in cubic inches) that Alina can make for Jen if she only has $30.00 to spend on materials? (Round your answer to three decimal places.)

[quartz gull]

I can’t figure this out

[fossil granite]

whats 9 + 10?

[dark tartan]

how does that relate

[fossil granite]

its a question

[dark tartan]

19

[quartz gull]

<@&286206848099549185>

[dark tartan]

AND why are you asking qn

[fossil granite]

sorry my calculator said 21

[dark tartan]

what

[quartz gull]

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[fossil granite]

nvm thank you for the help

[dark tartan]

💀

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[boreal heath]

I forgot how to do the substitution method

[marsh citrus]

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[tall pike]

17x^6 + 73x - 315 / 0,2 - x^7 = 0

[abstract knot]

!xy

[marsh citrus]

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.

@tall pike Has your question been resolved?

[marsh citrus]

@tall pike Has your question been resolved?

[tall pike]

!close

!stop

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[tall pike]

Yes

[sharp harness]

.close

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[fervent vale]

Hello what would be correct if they ask me for the first derivative

Should I simply it by the double angle identity

[lethal bridge]

both are correct

[fervent vale]

And when they ask me to find a tangent in a point 0, then still it doesn’t matter?

[lethal bridge]

simplifying things makes easier sure but they have the same meaning here

[fervent vale]

Thank you!

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[runic bone]

guys small question
since for charged condutors, electric field lines are normal to the sufraceat every point
we know that electric field lines can curve as well
so if a field line is normal to the surface then bends, is it still perpendicular to that surface?

[runic bone]

.

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[arctic thorn]

Can I say this diverges because if the harmonic series or is it not enough? Do I have to use integral test?

[nova totem]

Harmonic is fine

[arctic thorn]

ok thanks

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[mental hull]

Can a true patriot fact check this

[still temple]

I think this should be cos^2(x)-sinx cosx?

[mental hull]

Which?

[still temple]

the numerator since I assume you multiplied the numerator and denominator by cosx -sinx yes?

[mental hull]

yes, I was gonna say that I forgot to write down paranthesis but I did do cos^2x-sinxcosx

[still temple]

ah ok you just forgot it there

[mental hull]

yip

does the rest seem correct

[still temple]

I think something fucked up here

[mental hull]

why? I took a factor of 1/2 out because of the double angle

1/2(cos(2x)+1)-1/2sin(2x)

[still temple]

ah yes I misread it as sin(x)

then yeah I guess it's right

[mental hull]

yipee thanks

wolfram is returning something else and hmmm

[still temple]

although I'll say this is under a bunch of domain restrictions that it is possible that the function won't have the same antiderivative

[mental hull]

in a calc 2 class, I am assuming everything is integrable and smooth

[still temple]

I mean it should turn out the same

[mental hull]

it's returning ln sin+cos

and I am trying to reach that but I am failing

[still temple]

can you show

[mental hull]

sure

actually fuck

I am late

need to go sorry

.close

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[sleek crater]

Am I a genius or stoopid ?

solving for x

??

is this right so far

[gleaming pecan]

thrd line is bad

[sleek crater]

I multiplied both sides by 1/3

whats wrong with that

[gleaming pecan]

but sin is in denominator

[sleek crater]

yes

[gleaming pecan]

and in third line

you did not

[sleek crater]

3/sin3

they cancel

[gleaming pecan]

$\frac{3}{sinx}-sinx=2\Leftrightarrow \frac{1}{sinx}-\frac{sinx}{3}=\frac{2}{3}$

[elfin berry]

Joanna Angel

[gleaming pecan]

can you see it ?

[sleek crater]

oh yea

[gleaming pecan]

now, multiply by 3sinx

and you get euqations like quadratic one

$\frac{3}{sinx}-sinx=2\Leftrightarrow \frac{1}{sinx}-\frac{sinx}{3}=\frac{2}{3}\\\Leftrightarrow 3-sin^{2}x=2sinx\Leftrightarrow sin^{2}x+2sinx-3=0$

[elfin berry]

Joanna Angel

[gleaming pecan]

then you can substitute: u = sinx etc

[sleek crater]

CN you explain this substitution?

[gleaming pecan]

ok when you use new vriable u

then you get a quadratic equation like:

u^2 + 2u - 3 = 0

you know how to solve it ?

then, you can factor it, or use the delta formula

[sleek crater]

gotcha

[gleaming pecan]

cool

[sleek crater]

how did you know to multiply by 3sin at the top

seperate question ^ cant tell if I did smth wrong

[gleaming pecan]

it was oobvious fo rme, since, 3 and sinx were both in denomiantors

your new equation is sinx + 1 = cosx ?

[sleek crater]

yes

[gleaming pecan]

what did you do in second line ?

[sleek crater]

squared

[gleaming pecan]

it is wrong

[sleek crater]

why

[gleaming pecan]

first all you shud us eparenthesis

and second

yo do not know

i fboth sides are positive

so no

it wud be denmied by teacher

[sleek crater]

its recommend by the teacher

[gleaming pecan]

but you did not use

parenthesis

[sleek crater]

whats wrong if they are netative

-2^2 is still positive

[gleaming pecan]

look, wait =

$\left( sinx+1 \right)^{2}=cos^{2}x\Leftrightarrow sin^{2}x+2sinx+1=1-sin^{2}x\\\Leftrightarrow 2sin^{2}x+2sinx=0\Leftrightarrow sinx\left( sinx+1 \right)=0$

[elfin berry]

Joanna Angel

[gleaming pecan]

bu tperosnally i wud negate t my students

since, it requires chekcing

in the orignal

anyway

you forgot parenthesis

finally you got :

sinx = 0 or sinx = -1

[sleek crater]

Why you make me hate math

@gleaming pecan so are you saying

because we don’t know if its positive

we need to plug it back into equation ?

[gleaming pecan]

yes

[sleek crater]

isnt it bc squaring removes solutions

or adds extra solutions

so we need to check which is valid

what does it have to do with positive

[gleaming pecan]

sometimes squaring adds , how to say , foreign roots

flase roots

yes extra rotos you said

i say : obce pierwiaskti 🙂 in mhy native

so

i prefer to not deal with them

so if you sqaure, you must be caerful

itis even more important if it coem sto inequalities

but you have got onyl equations so far

that is another way :

$sinx+1=cosx\Leftrightarrow sinx-sin\left( \frac{\pi}{2}-x \right)=-1\\2cos\frac{\pi}{4}sin\frac{2x-\frac{\pi}{2}}{2}=-1\Leftrightarrow \sqrt{2}sin\left( x-\frac{\pi}{4} \right)=-1\\sin\left( x-\frac{\pi}{4} \right)=-\frac{\sqrt{2}}{2}\Leftrightarrow \\x-\frac{\pi}{4}=-\frac{\pi}{4}+2k\pi\text{ }\text{ }\vee \text{ }\text{ }x-\frac{\pi}{4}=\frac{5\pi}{4}+2k\pi\\x=2k\pi\text{ }\text{ }\vee \text{ }\text{ }x=\frac{3}{2}\pi+2k\pi$

[elfin berry]

Joanna Angel

[gleaming pecan]

I used formula of the form:

$sin\alpha-sin\beta=2cos\frac{\alpha+\beta}{2}sin\frac{\alpha-\beta}{2}$

[elfin berry]

Joanna Angel

[gleaming pecan]

my solution does not require checking original equation, if all is calcualted correctly.

[marsh citrus]

@sleek crater Has your question been resolved?

[sleek crater]

sorry had to go cuz driving but thank you so much

[gleaming pecan]

yw 🙂

[marsh citrus]

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[violet bane]

ugh does anyone know how to prove this?

[violet bane]

I tried using an example function where F(x,y) = f/g and finding the gradient of that but I get it in vector form idk how to put it in one thing like this

[sweet pawn]

what are the domain and codomain of f and g

[sinful skiff]

Well this form is a vector as well

[violet bane]

idk man ngl

[sinful skiff]

The numerator is a vector at least and the /g^2 can be viewed as a single term so it can be factored from the vector

[violet bane]

hmm yeah that's fair that makes sense, idk how to get it in that form tho I couldn't rearrange the terms after I did it

[sinful skiff]

Try to extract a 1/g^2 from both the terms

[violet bane]

yeah I did that

[sinful skiff]

Then the gradient of f and g should be (df/dx df/dy) and (dg/dx dg/dy) assuming f and g are both two dimensions

But I guess it’s also not specified what dimension f and g are

Also I’m using d but I’m referring to partial derivatives

[violet bane]

huh so I let F(x,y) = f(x,y)/g(x,y) and then I found Fx and Fy and let those be the components of the gradient

[sinful skiff]

Yeah that sounds right

[violet bane]

but I got <fxg - fgx, fyg-fgy> /g^2

[sinful skiff]

Try to turn that into the differences of two vectors now

The numerator

Break that single vector into two in a way that makes sense to you

[violet bane]

oh wait shit I see what you mean

that works

fuck thanks

[sinful skiff]

Also did you remember the chain rule btw

[violet bane]

uhh the chain rule

I did not lmao

where do I need the chain rule here

[sinful skiff]

Like you need to have f’ and g’ somewhere in there I’m pretty sure

Unless you meant to add the primes and forgot

Or I guess the prime notation is bad since f and g are multivariable

Oh wait that’s what you meant by fx and gx

The partials

Ok I see now

[violet bane]

yaa

I got it now thanks

it worked

wait so

[sinful skiff]

Nice

[violet bane]

how do I prove that it works for any number of variables?

do I just say the result holds for any number of variables or

[sinful skiff]

Well I guess for infinite dimension it would be different but for n dimensions you can just make them functions of n variables

X1, x2… xn

Then do the exact same thing

[violet bane]

o I see that's smart thanks

[sinful skiff]

I don’t know exactly how to do it with arbitrary or infinite dimensions

But I assume that finite dimension is almost expected

Cause you’ve probably never worked with infinite dimensions in calc 2 or even calc 3

Like most likely you are just working with functions from Rn to Rn

In fact as far as I know the gradient is only defined on finite dimensional spaces

[violet bane]

huh so I said F(x1, x2, ..., xn) = f/g and then found Fxn (partial derivative of F with respect to xn)

[sinful skiff]

You get something else when talking about banach spaces in general I think it’s called the frechet derivative

[violet bane]

not sure if that's enough but ig it works ?

[sinful skiff]

Well the gradient of F would require you to have a partial derivative in all the dimensions

[violet bane]

for a function of n variables I wasn't sure how to show that besides taking Fxn tbh

I can't exactly take Fx1 then Fx2 etc etc

[sinful skiff]

Yeah it becomes more cumbersome

I almost wonder if you can perform induction on the number of dimensions

Do you have product rule for gradient?

Like if that’s a given then you could also try using that

Actually I think another easier approach would be to prove that the quotient rule applies to any given coordinate of the vector

Like consider x_i

And the i-th component of the gradient

If you show that the I-th component of the gradient of f/g is equal to the i-th component of the vector on the right then you are also fine

And in that way your method of using xn is fine just to be pedantic I would use a different variable than n

Cause essentially you just want to argue component wise that they are equal

Definitely don’t use induction that wouldn’t help

[marsh citrus]

@violet bane Has your question been resolved?

[marsh citrus]

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[vocal cloud]

i have to calculate cardinal function for this data sets

[vocal cloud]

this is the formula

but i honestly forgot how to do it

this is what I think we have to do

l_o = (x-x1)/(x_0 -x1)*(x-x2)/(x_o-x2)

L_1 = (x-x_o)/(x_1-x_o) * (x-x2)/(x_1-x_2)

L_2 = (x-x_o)/(x_2-x_o)*(x-x_1)/(x_2-x_1)

is this right

[marsh citrus]

@vocal cloud Has your question been resolved?

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[grim rune]

Given a geometric progression (an) has a1 = 3 and 20a1 - 10a2 + a3 reach the smallest value. Find the 6th number of this progression

[proven vault]

try writing whole expression in terms of a2

[grim rune]

Uh

[proven vault]

a1 = a2/r and a3 = a2*r

[grim rune]

a1 = a2/q
a3 = a2.q

?

Ohhhhhh

[proven vault]

Idk if this will help or not but atleast try

[grim rune]

Ok

a2 ( 20/q + q - 10)

Right?

[proven vault]

Yes

[grim rune]

And now I got no clue

Lemme see

do I have to make the a2 = a1.q now?

Nvm

[grim rune]

Doesn't seem to help

[proven vault]

I think now you have to differentiate the function and find its minimum value

[grim rune]

What is "differentiate the function"?...

[proven vault]

dy/dx

[grim rune]

Oh wait nvm I searched it

Haven't learned it yet..

[proven vault]

Hmmm

Write the expression in a1

It'll make easier

[grim rune]

Ok

a1(20+q²-10q)

[proven vault]

And just know that differentiation of x^n is (n)x^(n-1)

[proven vault]

It'll become a1(2q-10)

[grim rune]

Ohhhh

[proven vault]

Put it equals to zero and get the value of "q"

q=5

[proven vault]

Alternatively you can also find minimum value at vertex of this parabola

[grim rune]

Ohhhh

[grim rune]

Still don't really get it..

[proven vault]

Do you know how to find vertex of a parabola

[grim rune]

a1(20+2q-10q??)

[proven vault]

Constant will become zero

Its 20x⁰

[grim rune]

Oh

[grim rune]

I got it now

Ty ty

So

a6 = 9375?

[proven vault]

Maybe

[grim rune]

Ty ty

.close

[marsh citrus]

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[violet stag]

this is just c and d right?

[marsh peak]

Yeah

[violet stag]

awesome

my notes define it in a pretty wordy way

just wanted to make sure

ty

.close

[marsh citrus]

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[molten tulip]

$\int\sin x\cos x\ dx$

[elfin berry]

Hazwi

[molten tulip]

when i do u sub for u = sinx i get 1/2 * sin²x+c

but when i do it for u = cosx i get -1/2 * cos²x +c

[frosty tundra]

C matters

if that's what you were going to ask

[molten tulip]

huh

care to elaborate?

[frosty tundra]

were you going to ask whether they're equal or not?

[molten tulip]

no i'm just asking how that is possible

[frosty tundra]

oh

how is what possible?

[molten tulip]

like getting a different answer for the same integral depending on how you choose "u"

[frosty tundra]

1/2 (sin^2(x)) + C is also equal to 1/2 (1-cos^2(x) + C aka 1/2 - cos^2(x)/2 + C

and that 1/2 gets "absorbed" with the C

[molten tulip]

ooooh alright

cool

ty

.close

[marsh citrus]

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[frosty tundra]

[marsh citrus]

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[foggy atlas]

Hey what properties does the relation {(a,b):a,b∈ℤ|a≡b(mod5)} have on ℤ?

It could have reflexive, transitive and symmetrical or only one of them, or none of them, or two, I'm not sure.

I'm a bit unfamiliar with mod, maybe someone can help me out

[bold ice]

$a \equiv b \mod 5 \iff 5 | a-b$

[elfin berry]

tales

[bold ice]

So we know $ 5 |0 = a-a$, so $a\equiv a \mod 5$

[elfin berry]

tales

[foggy atlas]

So I have this rule for reflexive right?

a≡a(mod5)

But isn't it not reflexive for a = 6?

[bold ice]

you mean $a \equiv a \mod 6$?

[elfin berry]

tales

[foggy atlas]

Oh no, hmm

So how can I know that this rule a≡a(mod5) is true for the relation?

I thought I can put in a value for a and test it but that's probably wrong?

[bold ice]

Is this the definition you are using?

[foggy atlas]

Oh nope, what is this definition?

What means 5|a-b?

[bold ice]

What is the definition of $a\equiv b \mod 5$ that you are using

[elfin berry]

tales

[foggy atlas]

"means that each element in the set is related to itself. In this case it means that for every a in ℤ it holds that a≡a(mod5)."

[bold ice]

No

[foggy atlas]

okay is the definition wrong?

[bold ice]

That is the definition of a reflexive relation

[foggy atlas]

Yea I want to know what the properties are from the relation and I wanted to start with if it's reflexive or not

[bold ice]

You have to show that $a\equiv a \mod 5$

[elfin berry]

tales

[bold ice]

But first, you have to know what $a\equiv b \mod 5$ means

[elfin berry]

tales

[foggy atlas]

Is it a mod 5 = b?

[bold ice]

'-'

[foggy atlas]

Sorry 😂

[bold ice]

https://www.youtube.com/watch?app=desktop&v=Q_V_itu_kbs

watch, then come back

[foggy atlas]

Okay what I've understood so far, a = b (mod 5) is a - b = 5

So a = a (mod 5) is a - a = 5. In the set of the relation are values for a and b, for example a = -10 and b = -5

So it's -10 - (-5) = -5, which is not 5 so it's not reflexive?

Or what is wrong in my calculation?

[bold ice]

'-'

Look at 0:47 on the video

[bold ice]

your definition is wrong

[foggy atlas]

5/(a-b)?

[bold ice]

$\dfrac{5}{a-b}$?

[elfin berry]

tales

[marsh citrus]

@foggy atlas Has your question been resolved?

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[stable crane]

why is the derivative of x^2 = 2x

[stable crane]

like I get the power rule or whatever you call it

[still temple]

okay

[stable crane]

but surely (d(x^2))/dx = dx/d

[still temple]

no

the 2 goes behind the x

its just power rule

[stable crane]

yea

[still temple]

nothing more

[stable crane]

but why does that work

like surely the gradient would just be divide function by x

like 7x has gradient of 7

and x^2/x = x^2-1 = x^1 = x

[still temple]

the power rule is based on the limit definition of the derivative

[stable crane]

gimme one sec to just catch up

[still temple]

but you are derivating it

you are not dividing it

[stable crane]

I'm pretty slow so gimma a min

so you're telling me

you can work out derivatives

using limits?

no that can't be it

[still temple]

im telling you that the derivative of x^n is given by limits

nothing more

nothin less

you asked me why it is like that

well that's why

[stable crane]

oh ok

ty

while your here

[still temple]

okay

[stable crane]

can you help me with something else

[still temple]

hopefully

[stable crane]

I'm still new to like calculus and all that so what is this

[proud ice]

The limit definition of a derivative

[stable crane]

ohhh

[proud ice]

Emphasis on definition

[stable crane]

why?

can you gimme one to practise as well

[proud ice]

Because it is the limit of the secant line between two points on the curve

It's basically slope approximation (rise over run) in the limit that run (given by the variable h) approaches 0

[stable crane]

ok several questions because I truly am horrible at this

what is the secant line

is that like the function

sec(x)

[proud ice]

Derivative of f(x)=x^2 is the best example to start with

[stable crane]

ok

[proud ice]

Secant line is the line between two points on a curve.

[stable crane]

so I replace the x with x + h

oh ok

nothing to do with sec(x)?

[proud ice]

Correct. Nothing at all

[stable crane]

hmm ok

anyways microsoft paint time for this

[still temple]

pure mathematical amazement

[stable crane]

lol

I never learned this

[proud ice]

Are you in a calculus class?

[stable crane]

no I just try to learn for fun : (

I am too small for calc classes

[proud ice]

Oh. Well good job making it this far but you kinda glossed over the most fundamental concept in all of calculus. Kind of what the whole thing is based on.

[stable crane]

yea my online friends kinda showed me it once

and then just went on about derivative of constant = 0

then derivative of ax = a

then he taught me power rule

[proud ice]

In your words, what does a derivative of a function give you? What is its purpose?

[stable crane]

the rate of change

[proud ice]

Very close

[stable crane]

at given x

as d gets closer to 0

[proud ice]

The instantaneous rate of change at a given x.

[stable crane]

difference being?

[proud ice]

Same difference between a secant line and a tangent line on a curve

[stable crane]

ohh

one is for two points and one is for two infinetely close?

so they are one

[proud ice]

You can have an approximation of the rate of change at x by picking two points close to x. That's basically the slope of the secant line of those two points. But using limits to get the tangent line precisely at x will be the instantaneous rate of change

[proud ice]

Yeah. Pretty much right

[stable crane]

ok

back to that derivative of x^2 using this new limit thing

so your saying

[proud ice]

The derivative is the limit to a perfect approximation of rate of change

[stable crane]

I can do any derivative with this

[proud ice]

Yes

[stable crane]

OMG

wow

[proud ice]

Wow indeed

[stable crane]

so let me try the x^2 for now

my handwriting bad on trackpad

[proud ice]

Some derivatives are harder than others though and you'll need to learn some techniques, but x^2 is a nice and easy start

[stable crane]

but is this good

so I replace f(x+h) now

this?

with lim obv

no way

tysm

[proud ice]

Ayy

Good job

[stable crane]

that was exciting ngl

[proud ice]

Early calculus learner here

[stable crane]

aw thanks

[proud ice]

You can try x^3 and x^4 if you like. Maybe find a pattern

sqrt(x) would also be a good one

[stable crane]

I know the power rule, I just didn't see why it was 2x and not x

although I still learned this rule

so that's even better than the power rule

[proud ice]

What math are you taking right now? You understand functions very well. I feel like you must be in trig or precalc

[stable crane]

nah

I'm a lot younger

I don't know how your school system works

[proud ice]

You can try proving power rule using definition of limit, but it's not obvious

What math are you in?

What's your class learning?

[stable crane]

um

factorization, but like everyone finds it easy

13 years old is my age

[proud ice]

Just made the cutoff for discord haha

[stable crane]

yea

ik

I broke the rules a bit but joined this server at 13, also started using this a lot more at 13

[proud ice]

Well you have a great understanding of functions. A lot of people really struggle with the notation

[stable crane]

thanks

[proud ice]

If you know geometry and trig, you really should be in a more advanced class

[stable crane]

yea we don't have classes yet

[proud ice]

But that depends on how well you know the material they plan to teach you know

[stable crane]

we in the uk have to wait till like 14/15 to do some exams

[proud ice]

Oh that's a shame.

[stable crane]

yea ik

[proud ice]

Keep studying then. You're doing great

[stable crane]

more time to prep ig

thanks

[proud ice]

Ping me if you ever need help. I respect your talent and hope you continue to enjoy math.

[stable crane]

thanks

that really inspires me : )

Everyone says such nice things

[proud ice]

Best server

[stable crane]

I hope I can persue maths

when I grow

because I love the subject

[proud ice]

You have an obvious skill at it.

[stable crane]

thanks

I just love the beauty of it at the same time tho

like infinite series and all that

and how there are still so many mysteries

[proud ice]

The more you learn, the more you will realize how little you know

It's a crazy feeling

[stable crane]

yea

I thought I had reached the edge of maths when I was like 10

[proud ice]

Same. Soon as I learned division and square root, I thought that was it. That was every button on my calculator.

[stable crane]

haha

you prob know a lot more that I even know exist

anyways I'll close now

[proud ice]

Yes. But I'm older.

[stable crane]

yea, exactly

[proud ice]

You learned calculus earlier than I did. Your growth beats mine

[stable crane]

thanks

[proud ice]

"I can see further now that I stand on the shoulders of giants "

[stable crane]

that's powerful

I'll remember that

.close

[marsh citrus]

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[still temple]

help. i dont know how to get the equation.

[paper raptor]

do you know how to get slope?

[still temple]

yes. the slope is 1/3

@paper raptor

[paper raptor]

and do you know how to get the y-intercept?

[still temple]

yes. it is 4

[paper raptor]

then you have the equation of the line:) $y=mx+b$

[elfin berry]

PajamaMamaLlama

[paper raptor]

where m = slope and b = y-intercept

[still temple]

i caan get it in y=mx+b but it wants it in point-slope form.

its y=1/3x+4

but thats not point slope form

@paper raptor

[paper raptor]

oh

well that's easy

can you find a point that the line goes through?

[still temple]

@paper raptor how abour 0,4

[paper raptor]

(you don't need to ping every time) and that works plenty well

[still temple]

sorry

[paper raptor]

now, point-slope form is as follows: $y-y_1=m(x-x_1)$, plug in and bob's your uncle :)

[elfin berry]

PajamaMamaLlama

[still temple]

so its y-4=1/3(x-0)?

[paper raptor]

yep!

[still temple]

thanks

[boreal rose]

who is bob

[still temple]

his uncle

[paper raptor]

https://en.wikipedia.org/wiki/Bob's_your_uncle

it's an idiom

[boreal rose]

ohhhh average UK ppl

that's why US >>

[paper raptor]

not UK :), just a fun expression to say

[still temple]

im british also

[boreal rose]

well I'm sorry about that

[still temple]

your sorry im british?

[boreal rose]

mhm

[still temple]

but we are better then the US. we got W health care and W free schooling

where is your free healthcare.

hmmmmmmm....

[boreal rose]

but do you guys have joe biden

yeah....

[still temple]

you just insulted yourself 💀

imagine

[boreal rose]

what's wrong with good old joe

[still temple]

https://tenor.com/view/biden-joe-biden-american-president-potus-biden's-fall-gif-2907581033668205349

the leader of your country

[old glacier]

not sleepy joe

[still temple]

https://tenor.com/view/uk-boris-johnson-parliament-uk-government-bojo-gif-26174772

[old glacier]

who relies on chocolate icecream every sunday

[still temple]

He's a Trump fan, trying to look like Trump too

[old glacier]

mf needs a babysitter

[still temple]

(y = \frac{1}{4}x + \frac{7}{4} - 1), (y = \frac{1}{4}x + \frac{7}{4} - \frac{4}{4}), (y = \frac{1}{4}x + \frac{3}{4}).

$(y = \frac{1}{4}x + \frac{7}{4} - 1), (y = \frac{1}{4}x + \frac{7}{4} - \frac{4}{4}), (y = \frac{1}{4}x + \frac{3}{4}).

$(y = \frac{1}{4}x + \frac{7}{4} - 1), (y = \frac{1}{4}x + \frac{7}{4} - \frac{4}{4}), (y = \frac{1}{4}x + \frac{3}{4}).$

[elfin berry]

Illumiant

[boreal rose]

NOT BORIS JOHNSON

GRACE

[still temple]

$[y = \frac{1}{4}x + \frac{3}{4}]$

[elfin berry]

Illumiant

[marsh citrus]

@still temple Has your question been resolved?

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[stable crane]

how do you prove the power rule for derivatives?

[stable crane]

I tried and failed with this new limit thing I learned

[native frigate]

limit definition of the derivative should work iirc

can you show your work

[stable crane]

oh

it very basic

but sure

I just lerned limit

the thinks that look like 9's are a's

i think that right

[native frigate]

that doesnt seem right

[stable crane]

is it not?

for f(x) = x^a

[native frigate]

wait i think i misread

is it your assignment to prove this, or are you doing it for yourself

the proof is less trivial than I thought so if its for class there might be other useful context

[stable crane]

oh ye

sorry there's meant to me a plus in (x-h)

no

like just the power rule

(x^a)' =

whatever

[native frigate]

but it is a class assignment to prove why this works?

[stable crane]

no

for fun

[native frigate]

do you know implicit differentiation?

or the binomial theorem

[stable crane]

I wish I could say yes, but I am super new to calculus

oh

binomial theorum yes

i think

maybe not actuakky

[native frigate]

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-diff-1-optional/v/proof-d-dx-x-n

[stable crane]

yea

[native frigate]

i didnt watch the video, but this is the binomial theorem version of the proof

you should be able to follow

[stable crane]

what are the delta x for

is that just my h

[native frigate]

yes

[stable crane]

ok

[native frigate]

$\delta$ means "change in" usually

[native frigate]

$\Delta$

[native frigate]

so its just a small change in x, which is what h also is

[stable crane]

I thought that was just d

or maybe it just another way to write d

[native frigate]

i used the wrong delta in my first message, the triangle one is the one i want

[stable crane]

oh

[native frigate]

$\delta$ is also for partial derivatives

[stable crane]

capital delta

what is a partial derivative

[native frigate]

so you know how by default you have y = f(x)? When you take the derivative you are taking it with respect to x (change in slope on the x axis)

but you dont have to do that

if you had some function like f(x, y) = xy, you could take the derivative with either respect to x or y

[stable crane]

hmmm

[native frigate]

which are called partial derivatives

[stable crane]

ok

I a bit confused but it's fine

[native frigate]

probably can find better resources on khanacademy or pauls online notes

although if you are just doing power rule, I dont think you should skip to it

iirc its a calc2 topic, and you would be in calc1 or precalc

[stable crane]

I don't do calc, I just learn for fun

so I don't know what I am

I know most of precalc I think

[native frigate]

well, i would definitely take a look at that khanacademy video i linked, its the standard binomial theorem proof of the power rule

any other questions?

[stable crane]

no

I think I got i

it*

ty

.close

[marsh citrus]

Channel closed

Closed by @stable crane

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[vast jolt]

i need help to undertsand something about transposition. I read that this formula can be a triangle.

 pl
-----
R x A 

and i can rearrange that to get what i need. what i don't understand it why the example above moves the A left first

what can't it move the pl over right away? the rules of transposition say it must be the opposite ? so wouldn't it become

R
-- = A
pl

[marsh citrus]

@vast jolt Has your question been resolved?

[marsh citrus]

@vast jolt Has your question been resolved?

[real shore]

@vast jolt

still here?

your example explains in detail the process of solving for a certain variable

the triangle method is a quick one used when you need to do stuff fast

the actual rule for transposition are as shown

usually we keep the variable we solve for left

the example moves A from right bottom to left top

so you would have RA/1 = ρ1/1

We can write anything as a fraction by putting 1 below

1 = 1/1

2 = 2/1

167 = 167/1

[real shore]

then he divides with the assembler of the unknown Value

which would be R

he divides both sides

we do this to ''get rid'' of the assembler

[vast jolt]

but what i don't understand, is if i use the triangle rule, i can simply always move the top over withouth changing it.

and get the correct answer

[real shore]

.

the official explanation behind that rule is the steps used in the example

for example if you were in a formal examination that required you to reason your course of action

you couldnt just say the triangle method

[vast jolt]

why do the move A first? do they always move the denominator, get rid of that, first?

[real shore]

we move A because we want to solve for A

[real shore]

at the top here it says ''to find a formula for A''

If you wanted a formula for ρ1 you'd move ρ1

[vast jolt]

ok, but why is it not allowed to just move the pl over when that is going to bve correct?

[real shore]

why would you move ρ1?

[vast jolt]

because that would give you the correct answer

pl
-- = A
R

[real shore]

that is incorrect

because if you move ρ1 you would have

ρ1 1
-- = --
R A

because A is at the bottom

[vast jolt]

the triangle shows that pl will never be divided by anything. so i am confused why this "quick" method somehow violates the rules of transposition?

but gives the correct answer

[real shore]

it doesnt violate any rules

but it is not like

a formal reasoning

if an exercise asks you to use the transporation rules

you'd have to use these steps

[vast jolt]

that i can understand and i'd not question that.

i'm just cannot seem to see how transpostion takes a weird route to a solution the trianlge says is evident.

as far as i can see, in any example of a triangle formula, you never divide the top and just move it.

and you get the correct solution?

[real shore]

okay

wait

suppose this is your formula

a = b/c

[vast jolt]

ok one sec

wait a sec

[real shore]

yes

[vast jolt]

 b
----
a x c

[real shore]

you want to do that?

with the formula?

[vast jolt]

that is my triangle

[real shore]

wait

Does this help?

This better

[vast jolt]

so to get c, i move b over

a = b/c > b/a = c

[real shore]

yes

c = b/a

[vast jolt]

but that transposition says b would be opposite?

a/b = c

right?

[real shore]

wdym opposite

[vast jolt]

the rule is if you go over the = it swaps place or -/x///+

so it would go under a?

i'll show the whole thing

[real shore]

okay

generally

[real shore]

ohh

and you wanna know

why the symbols dont change?

when the letters cross the = ?

[vast jolt]

for b i this example.

or why they don't move b but go the long way around.

[real shore]

okay