help-13

[hollow osprey]

but this isn't a test right?

[light fern]

nah this is a practice test

this is basically a study guide

test is tomorrow

[hollow osprey]

alright

[hollow osprey]

no answer key?

[light fern]

Can you help me with this one too?

[light fern]

Maybe in google classroom, one se

Sec

[hollow osprey]

yep okay so first i want you to consider

use FTC part 2 to write this integral as a difference of the antiderivative evaluated at separate values (2x and 1)

then you can use the chain rule to differentiate it

[light fern]

Whats FTC?

[hollow osprey]

fundemental theorem of calculus

[light fern]

(Also there is an answer key but I want to solve with minimal help)

Alright

[hollow osprey]

this one

$\int_{a}^{b} f(x) \dd x = F(b) - F(a)$

[wraith dagger]

Yottachad

[light fern]

Like this?

[hollow osprey]

not quite

that wouldn't be the antiderivative

oh crap

[gaunt hamlet]

The question says F'(x), where F(x) is the integral

[hollow osprey]

it already uses capital F sorry for the confusion

[gaunt hamlet]

You only need part 1

[light fern]

so i only need the part with 2x^3?

[gaunt hamlet]

I mean part 1 of FTC

[hollow osprey]

yeah but its easier to understand if you use part 2 first

[light fern]

is part 1 F'(x)=f(x)?

[hollow osprey]

so michael im gonna call it lower case g instead

so we have:

[light fern]

alright

[hollow osprey]

$g(x) = \int_{1}^{2x} \frac{1}{1+t^3} \dd t$

[wraith dagger]

Yottachad

[gaunt hamlet]

I guess that's true since the upper bound is 2x

[hollow osprey]

and we wanna find g'(x)

[light fern]

and that would be 1/(1+t^3)

but since the upper bound is a function itself

we do something different right>

?

[hollow osprey]

yeah pretty much you have to use chain rule

[light fern]

am i taking antiderative or derative?

[hollow osprey]

if you call the antiderivative o fthis

= F(t)

ill write thiso ne out

$F(t) = \int \frac{1}{1+t^3}dt$

[wraith dagger]

Yottachad

[hollow osprey]

we know now that we can write g instead as:

$g(x) = F(2x) - F(1)$

[wraith dagger]

Yottachad

[hollow osprey]

by FTC part 2

[light fern]

Alright

[hollow osprey]

so from here

you should just take the derivative of both sides

you get:

$g'(x) = F'(2x)\cdot \dv{x} (2x)$

[wraith dagger]

Yottachad

[hollow osprey]

note that F(1) is a constant

and by FTC part 1

$F'(x) = \frac{1}{1+x^3}$

[wraith dagger]

Yottachad

[hollow osprey]

so now you just have to plug in values to find the answer

this was a trash explanation here im gonna try to find a vid

https://www.youtube.com/watch?v=aeB5BWY0RlE

we ❤️ organic chemistry tutor

[light fern]

Yeah hahah, i watched some vids of him

Can you timestamp a good spot?

[hollow osprey]

yeah hold on

5:05

that's what i was trying to get to but i did it badly

[light fern]

from watching that part i got 1/(1+2x^3) times 2

$F'(x) = \frac{1}{1+2x^3} {2}$

[wraith dagger]

MichaelMichaelMicha

[light fern]

that doesnt match any of the answers tho

oh waittt

It would be (2x)^3

so the 3 would go to the 2 right?

[light fern]

So basically for a problem like this, do i basically ignore the constant and evaluate the function in the upper limit?

What if the function is in the lower limit? Do i do the same or should i flip the integral?

[hollow osprey]

flip the integral

or you could write out what i was doing

this stuff

$\int_{x}^{b} g(x) \dd x = G(b) - G(x)$

so then you could see how when you take the derivative you would get the negative version

[light fern]

alright

[wraith dagger]

Yottachad

[light fern]

Can you send me a video where you learned how to do the fancy pictures? I want to learn how to do it

[hollow osprey]

i didn't watch a vid i just watched ppl do it in the #calculus

channel

but this is like almost all you will ever need

$\int_{a}^{b} f(x) \dd x$

[wraith dagger]

Yottachad

[hollow osprey]

$\dv[n]{f}{x} \text{ and } \dv{x}$

[wraith dagger]

Yottachad

[hollow osprey]

$\frac{1}{x+1}$

[wraith dagger]

Yottachad

[hollow osprey]

and most importalty

$\cdot \text{ for multiplication}$

[wraith dagger]

Yottachad

[hollow osprey]

do NOT use *

or you will get hit with the

[gaunt hamlet]

This server uses LaTeX. You can find a variety of tutorials online

[light fern]

Alright thanks for the help

[hollow osprey]

do they show the \dv[n]{f}{x}

i ddin't find that out for months

[hollow osprey]

no problem if you have no more questions you can use .close to close the channel

[gaunt hamlet]

No idea. My knowledge of LaTeX is solely from seeing people use it here lol

[light fern]

.close

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[steel canopy]

Is possible to have cross product as orthogonal?

[steel canopy]

I feel it should no because u dot product v in order to be orthogonal

I am pretty sure orthogonal represent perpendicular and cross product means parallel?

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[keen mulch]

I am stuck in this problem where we need to show that sqrt(2) always lie between a/b and (a + 2b)/(a + b) where a and b are positive integers.

[cedar kiln]

@keen mulch Has your question been resolved?

[keen mulch]

<@&286206848099549185>

[jagged parcel]

Hmm, I have a solution. Will explain after the photo loads.

,rotate

[jagged parcel]

Ok so, In the event that a/b is smaller than a + 2b / a+b
a/b is smaller than root 2 which is smaller than (a + 2b) / (a+b)

In the event that a/b is larger than (a + 2b) / (a + b)
Then a/b is larger than root 2.

I will keep working on this, as now you have to prove that in both these situations, the root 2 is smaller/larger than the a + 2b / a+b

[keen mulch]

okay

okay so as it turns out we can do a little manipulation which would help us prove it

@jagged parcel

[jagged parcel]

Ah cool! Yeah so with both in hand you can say that it has to be true, nice

[keen mulch]

👍

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[keen rapids]

What is the remainder when x^2-x-2 is divided by x-1?

[glad kestrel]

what have you done so far

[keen rapids]

tried to devide them

[glad kestrel]

do you have work

[keen rapids]

hold on a sec

x^2-x-2/x-1
=x+-2/x-1
assuming; (r) -2

I'm not sure

[glad kestrel]

do you know polynomial long division

[keen rapids]

yeah, I have one on paper but it's a lil messy

[dusty hazel]

That's fine.

[keen rapids]

want me to send the paper work?

[glad kestrel]

yeah

[keen rapids]

Give me a munite

Sorry if it took so long. I had to rewrite it into small pieces. It's still not clear though.

[glad kestrel]

looks good

[polar mist]

So that’s the remainder

[glad kestrel]

not the whole thing

[keen rapids]

-2 right?

[glad kestrel]

yea

[keen rapids]

yeah just had to make sure. Thank youu

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[turbid flax]

can someone explain to me why 5 divided by 1/4 is 20 , shouldn't it be below 5? and why is 2.5 divided by 1/2 =5? shouldn't it be 1.25? it's literally the most basic thing i just don't get it lol

[glad kestrel]

dividing by a fraction is the same as multiplying by the reciprocal

if you were to divide 5 by four, it would be smaller than 5, yeah

but you're dividing by 1/4

[turbid flax]

so kinda like 2 negatives=positive?

[glad kestrel]

well not the same concept but if you need something to relate it to sure lol

[turbid flax]

ok thx for the help

[cedar kiln]

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[zenith python]

What is the probability that Marco spends at least 240min on online gaming if his mean time spend on it is 279 min with a standard deviation time of 15min?
mean=270min
Standard deviation=15min
x>240min or x<240min?

[tropic oxide]

at least 240min

[zenith python]

So x>240?

[tropic oxide]

yes

[zenith python]

Okiii thankss ❤️

[tropic oxide]

at least means ≥ rather than > but for your purposes it doesn't make a difference

[zenith python]

Okay okay, tyy

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[coarse stream]

help with median

[cedar kiln]

@coarse stream Has your question been resolved?

[trail jackal]

The better question is: Why is there no student at all who attended school everyday? What a horrible school must this be... 🤣

With which part are you struggling?

[cedar kiln]

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[cunning nova]

Is the error calculated in the second equation equal to the error of the 'k'th neuron?

[cunning nova]

To be clear. This is the backpropagation process of an artificial neural network

Here is an implementation of the process in python, if that helps

[cedar kiln]

@cunning nova Has your question been resolved?

[cunning nova]

<@&286206848099549185>

[obsidian coral]

One, more context is probably going to be needed then the 2 formulas and works.
Two, that's very code related so you should ask in a coding server. Sure it has math but all I see is code, and it could be some logic issue in the code

[cunning nova]

Alright. I think I got it now. It's really hard to get help on that topic. It's way to much code to be reviewed as a whole but the network structure is also pretty individual, so it's hard to generalize mathematical equations or coded functions. It was worth a try

.close

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[lost thistle]

Can someone tell me how to solve this question

[gentle lintel]

$14 = 3^2 + (\sqrt{5})^2$

[wraith dagger]

Chromium

[lost thistle]

I didn't get it

[gentle lintel]

therefore $14 - 6\sqrt{5} = 3^2 + 2 (3) (\sqrt{5}) + (\sqrt{5})^2$

[wraith dagger]

Chromium

[gentle lintel]

this is the key to your problem

[lost thistle]

Im not getting it 😭

[gentle lintel]

do you not understand this formula?

or how i chose to rewrite it this way

[scenic heath]

What do you do to get rid of square roots?

[gentle lintel]

you.. square them??

[scenic heath]

That was more a question for SenZ not for myself lel

[lost thistle]

Yeah square them, but there is nothing in rhs, so we can't square right? Or can we?

[scenic heath]

Square of 0 is...?

Or at least that is my thinking

[lost thistle]

No i understood what u wrote, but i can't connect it to the question

[gentle lintel]

this is the question

[lost thistle]

It's 0 but there is no 0 in the question right

[gentle lintel]

do you know how to go after this?

[lost thistle]

I think I'm kinda getting it, we shud write everything in their square form so we can cancel the square root?

[gentle lintel]

yea

usually, for questions like $\sqrt{a + \sqrt{b}}$, the inside can be rewritten as $(x + \sqrt{y})^2$

[gentle lintel]

like here

[jaunty mural]

(x+sqrt y)^2

a+c sqrt b

[wraith dagger]

Chromium

[jaunty mural]

urgh

just put coefficients in places smh

[gentle lintel]

whatever, $(thing_1 + thing_2)^2$

[wraith dagger]

Chromium

[jaunty mural]

$$\sqrt{a + b\sqrt{c}} = \sqrt{(x + y\sqrt{c})^2}$$

[gentle lintel]

i guess?

[wraith dagger]

Shuri2060

[gentle lintel]

hope senz gets the point

[jaunty mural]

chromium is saying that usually they are created like this

[lost thistle]

Oh no my tiny brain is not being able to handle all this

[jaunty mural]

Take care to always take the positive root

[lost thistle]

Uhh I'm not understanding, can we start over please?

[jaunty mural]

You assume

the inside of your square root looks like this

So the first one, we have a = 14, b = -6, c = 5

[gentle lintel]

solve by assumption

[jaunty mural]

$$\sqrt{14 + -6\sqrt{5}} = \sqrt{(x + y\sqrt{5})^2}$$

[wraith dagger]

Shuri2060

[jaunty mural]

Now try to find x and y

You can do this by expanding the thing under the square root on the right hand side

Expand the thing under the square root

[lost thistle]

x² + 2xy root c + y² c

[lost thistle]

Isn't that x +y root c itself

[jaunty mural]

yes so now you can figure out what x and y must be

$$\sqrt{a + b\sqrt{c}} = \sqrt{(x + y\sqrt{c})^2} = \sqrt{(x^2+y^2c)+2xy\sqrt c}$$

[wraith dagger]

Shuri2060

[lost thistle]

How to find x and y? By equation with 14 -6 root 5 ?

[jaunty mural]

....

You are given a b and c

Can you see how these 2 must correspond

[gentle lintel]

(wait how did you blur the middle)

[jaunty mural]

ShareX

[lost thistle]

Ohhh so a = (x²+y²c)? And b = 2xy?

[jaunty mural]

yes

Once you have found an x and y that works

then you know it must be like this

And you take the positive root

It could be +-(x+y sqrt c) depending on which one is positive

[lost thistle]

Ahh I'm lost 😭 sorry

[jaunty mural]

What equations have you got.

[lost thistle]

Can we do with simple numbers, all these variables got me confused

[jaunty mural]

???

[jaunty mural]

You literally just wrote this down

$$a = x²+y²c$$
$$b = 2xy$$

There is nothing to be confused about, all these letters just represent numbers

You are given a, b, c

First do the innermost one

[wraith dagger]

Shuri2060

[lost thistle]

Ok so I need to write 14 as (x² + y² c)right

[jaunty mural]

What are a b and c???????

[lost thistle]

A is 14 b is 6 and c is 5?

[jaunty mural]

no.

$$\sqrt{14+6\sqrt5}$$

[wraith dagger]

Shuri2060

[jaunty mural]

.

[lost thistle]

Is that correct ?

[jaunty mural]

.

Does it look correct

14 + 6 sqrt 5

huh?

[lost thistle]

So a is 14 b is -6 and c is 5 ?

[jaunty mural]

yes

See how this matches up?

Now you want to solve the equations you got from here

for x and y. So plug a b c in

[lost thistle]

Ok so 14 = 3²+ (sqrt5)² ?

[jaunty mural]

no idea where you're getting this from

[lost thistle]

Sry my bad

[jaunty mural]

you literally just told me
a = 14
b = -6
c = 5

[lost thistle]

Ohhh i need to find x and y?

[jaunty mural]

That's what we've been saying for the last while...

Otherwise what the heck is the point of all this

[lost thistle]

Sry i just got it now, i was confused till now

[jaunty mural]

You plug in a, b, c and then try to find x and y which solve the equations

[lost thistle]

I got xy as -3

And x² +y² as 14/5

[jaunty mural]

no.

BODMAS

Since when does p + qr mean (p + q)r

p + qr means p + (qr)

[lost thistle]

Ohh sry sry i didn't notice it
So I got x²+5y²-14=0

[jaunty mural]

The first step to do is just plug the numbers in

no need for any fancy rearrangements

$$14 = x²+5y²$$
$$-6 = 2xy$$

[wraith dagger]

Shuri2060

[lost thistle]

14= x²+5y²

[jaunty mural]

Then like chromium suggested

try looking for integers which make this work (because the question is probably made this way)

If you can't, then you solve the simultaneous equations like normal

Notice that you have 2 equations and 2 variables, so it should be solvable.

[lost thistle]

Ohhhh ok ok i ll try solving

[lost thistle]

Oh wow now I'm having trouble solving this also 😭

[lost thistle]

I ll try doing that way

I got it

[jaunty mural]

Really???

[lost thistle]

x= 3 and y = 1 ?

[jaunty mural]

Well yes

I mean there really wasn't very many to guess

[jaunty mural]

Actually no, this is clearly wrong.

There are only a few things to guess, you shouldn't be having trouble.

[lost thistle]

Ohh sry y = -1

Is that correct?

[jaunty mural]

I'm not going to hold your hand all the way through

You should be able to check this yourself.

The process for the next steps are exactly the same as here.

[lost thistle]

Ok i understood now only that we trying to make it in the form a²-2ab+b² so that we can then later write it in (a-b)²

Sry i just got super confused when i saw so many variables

Thanks for ur time

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[plush walrus]

How to prove that there are infinitely many primes of type 4k+1
Without using "congruence". Because I don't have idea about it. This problem encountered me before Congruence chapter.

[plush walrus]

<@&286206848099549185>

[jaunty mural]

Doesn't stop you using a congruence argument regardless.

Congruence is literally just shorthand

The fact all integers being even or odd can be used before that chapter

Same for other remainders upon division.

[plush walrus]

I know how to prove that there are infinitely many primes of type 4k +3

But don't know how to prove for 4k+1

So can you help me?

[jaunty mural]

I don't know either.

Try to mimick your previous argument with some changes.

[plush walrus]

I have tried and searched in google. I am unable to find.

[jaunty mural]

contradiction

attempt to construct a 4k+1 not on your list.

[plush walrus]

Yes
But even number of 4k+3 type will lead to a 4k+1 type. So. How to show that there must be at least one factor of type 4k+1

Prime factor*

[cedar kiln]

@plush walrus Has your question been resolved?

[jaunty mural]

#elementary-number-theory

[gentle lintel]

please only tag helpers after 15 minutes

[cedar kiln]

@plush walrus Has your question been resolved?

[cedar kiln]

@plush walrus Has your question been resolved?

[cedar kiln]

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[celest saffron]

q1 = 12.0 C is located 10.0 m left of q2 = –15.0 C. What is the electric field 20.0 m below the midpoint of q1 and q2?

[celest saffron]

i have a question on how to find the electric field at 20.0 m

[cedar kiln]

@celest saffron Has your question been resolved?

[cedar kiln]

@celest saffron Has your question been resolved?

[upper garnet]

with the use of angles

then adding the horizontal part and vertical part of the field together

https://m.youtube.com/watch?v=FW5HHS1V8Ko

[celest saffron]

.close

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[celest saffron]

thanks btw ❤️

[cedar kiln]

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[cobalt gulch]

I need help with binomial distribution

[cobalt gulch]

I need to figure out how to figure out the possibility for at least 6 out of 10 possibilities

I wrote this but I believe this is for exactly 6 not at least 6

[exotic elk]

What was ur calculation to get to 0.143

[cobalt gulch]

It gave me the answer cause I was wrong

[exotic elk]

do you have the formula or do i need to type it

because 0.143 is correct

(as u said)

[cobalt gulch]

I belive the formula is this no?

[exotic elk]

$$f(x) = \binom{n}{x} p^x q^{n-x}$$

[wraith dagger]

MattDog_222

[exotic elk]

so $n=10$ because you are sampling 10 people. we want $x=5$, and we are given that our "Success" (people you dont believe in the newspapers) is $p = 0.66$

[wraith dagger]

MattDog_222

[cobalt gulch]

I figured out the one for 5 I’m trying to figure out how to find out if it’s at least 6 people

[exotic elk]

at least six is $P(X \leq 6) = 1 - P(X > 6) = 1 - \Big( P(X =7) + P(X=8) + P(X=9) + P(X=10) \Big)$

[wraith dagger]

MattDog_222

[exotic elk]

there's not really a shortcut other than using a calculator or using the one with the less sums

[cobalt gulch]

Ahh I get it now

[cobalt gulch]

For the last bit would it also include +P(x=6) since it’s equal to or greater than 6 but less than 10?

[cedar kiln]

@cobalt gulch Has your question been resolved?

[exotic elk]

sorry was eating. I misread it. i thought it said P(X<=6). P(X >= 6) includes the sum of P(X=6)

theres two ways to go about it still

$P(x \geq 6) = P(X = 6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)$

[wraith dagger]

MattDog_222

[exotic elk]

or
$P(x \geq 6) = 1 - P(X < 6) = 1 - \Big[ P(X = 1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)\Big]$

[wraith dagger]

MattDog_222

[exotic elk]

https://homepage.divms.uiowa.edu/~mbognar/applets/bin.html
u can also compute this with online calculators or a TI83/84 also has pbinom and cbinom

[cedar kiln]

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[shadow oar]

need help for c

i think i can solve d and e once i get c done

i suppose the weight of the boy is 400N because thats the normal force when the lift is on constant velocity?

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[meager reef]

How many possible musical arrangements can be formed in your preferred music? Present the arrangement using the fundamental counting principles. (MATH)

[meager reef]

im really confused about this by what they mean on "musical arrangements"

its kinda complicated

[cedar kiln]

@meager reef Has your question been resolved?

[meager reef]

<@&286206848099549185>

[wraith crypt]

well, without more context everyone here is also really confused by what your question means

[meager reef]

well, its like any songs and it shows here that i need to find "musical arrangements" that can be formed in that music and i don't even know how

ima ask my classmate ig

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[atomic cipher]

quick question - how do I solve this without a calculator

[bold vine]

Dividing by a fraction is the same as multiplying by the reciprocal

More generally:

a / (b/c) = (ac)/(b)

@atomic cipher

[atomic cipher]

like this?

[dim kayak]

(5.3)/1

[cedar kiln]

@atomic cipher Has your question been resolved?

[gentle lintel]

yep

[cedar kiln]

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[hallow stream]

what rule is this?

[bright surge]

express the average value as an integral first

[hallow stream]

what do you mean?

[hallow stream]

I thought this would be related to theorems in definite integrals

[bright surge]

yeah

but what does average value mean

[hallow stream]

is it f(b)-f(a)/b-a

[bright surge]

no

it's actually $\int_a^b \frac{f(x)}{b-a} \dd x$

[wraith dagger]

RYC for mod⁴ (meg)

[hallow stream]

so that is equal to 12

[bright surge]

yep

[hallow stream]

so how do they link together? I guess that b-a would be related to 8?

[bright surge]

uh

try substituting in the actual values

[hallow stream]

so 6

[bright surge]

yep!

[hallow stream]

12 = $\int_-2^4 \frac{f(x)}{6} \dd x$

oops

[bright surge]

yea

[hallow stream]

is this right?

[bright surge]

yes

[hallow stream]

so how do I solve for f(x)? Do I just integrate it as if it were X?

[bright surge]

no!!

pull the 1/6 out first

and then multiple by 1/8

or just multiply by 3/4

[hallow stream]

why?

[bright surge]

you want $\int_{-2}^4 \frac{f(x)}{8} \dd x$. you have $\int_{-2}^4 \frac{f(x)}{6} \dd x$. you also have $a\int f(x) \dd x = \int af(x) \dd x$.

[wraith dagger]

RYC for mod⁴ (meg)

[hallow stream]

im kind of confused

I dont know how the first one links up to the 2nd one

oh I think I get it

I have to make them equal

[cedar kiln]

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[stoic plinth]

Hey, I would like to know if the scalar product is this solution. I am not really familiar with the notion and we did not get a clear definition in class

[tropic oxide]

you have the definition written out right there and you applied it correctly and made no mistakes in the arithmetic.

[stoic plinth]

thanks, really appreciate your help!

[crimson sedge]

Yes, the top part is the definition for arbitrary numbers

[stoic plinth]

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[gaunt lake]

need help

[crimson sedge]

with?

[wraith crypt]

and what have you tried?

[gaunt lake]

uh ive tried expanding it

but i got confusded with the j and n

[gaunt lake]

with the qn i sent on how to expand the summation

[crimson sedge]

What do you get for j=1

[gaunt lake]

I only did 2 j squared which got me -2+2^n

[cedar kiln]

@gaunt lake Has your question been resolved?

[cedar kiln]

@gaunt lake Has your question been resolved?

[cedar kiln]

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[sterile bolt]

No.12

[sterile bolt]

<@&286206848099549185>

🗿

[sterile bolt]

<@&286206848099549185>

How do I do no.12

[cedar kiln]

@sterile bolt Has your question been resolved?

[sterile bolt]

Help

<@&286206848099549185>

[trail jackal]

use tan(A)=sin(A)/cos(A) and cos^2(A)+ sin^2(A)=1

[sterile bolt]

Where and when

[trail jackal]

start with tan(A)^4-tan(A)^2-1 and rearrange that with the above equalities until you get 0

[sterile bolt]

[trail jackal]

use the first equality, then bring everything on the common denominator and use the second one to red rid of all cos terms

[sterile bolt]

Idk what to do from here

[trail jackal]

use the second one to red rid of all cos terms

[sterile bolt]

Okay so I bring in all sin terms?

In place of the cos ones?

[trail jackal]

yes

[sterile bolt]

Ok

Oh geez

Phew

Thanks

Wish I could figure these out but I always get stuck on something in these trigonometry questions

Even though I know all the identities...

[sterile bolt]

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[azure plinth]

Hey guys. I'm just curious about how to determine the signage without graphing. I used a pythagorean identity to make tan^2(x) = sec^2(x) - 1. Not sure if it's negative just based on that idea alone.

[azure plinth]

Is there anything else I could be missing?

[glad kestrel]

what quadrant are you restricted to?

[azure plinth]

The second, based on the question.

[glad kestrel]

what is the sign of tangent in the 2nd quadrant?

[azure plinth]

negative

[glad kestrel]

and square it?

[azure plinth]

positive

[glad kestrel]

now secant in the 2nd quadrant?

[azure plinth]

negative

mmkay

I get that, yeah

so it's negative

[glad kestrel]

yep

[azure plinth]

Was my work valid though?

[glad kestrel]

yes

[azure plinth]

Cool. Thank you so much.

[glad kestrel]

👍

[azure plinth]

.close

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[gentle lintel]

where can i find examples for applications of taylor series online?

[gentle lintel]

limits, integrals, etc

[south tundra]

Taylor series are used for approximating some functions

For example, how do you think the calculator approximates values of sin and cos?

(or any other trig functions)

[gaunt hamlet]

I always assumed their code just asked God for the answer

[rough pebble]

yes and he uses the taylor series

[south tundra]

Correct

[jaunty mural]

applications? D:

[south tundra]

What's wrong tho

[gentle lintel]

these kind of applications..?

[jaunty mural]

We use it to find the Laurent series heheheh

heheheeheh

[gentle lintel]

clever ways to use taylor series, in general

are there related examples online

[jaunty mural]

Taylor -> Laurent, very clever

Ok sorry lmao

[jaunty mural]

Literally limits

You can stick in the taylor series

[gentle lintel]

like..?

[jaunty mural]

sinx/x

[gentle lintel]

bruh

[jaunty mural]

what.

you will notice everything vanishes except x/x = 1

im just giving a basic example of a limit where you can stick the series in

[fallen heath]

$\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}$

[gentle lintel]

tasty

[jaunty mural]

candy crush 🍬

[gentle lintel]

d i v i n e !

[gentle lintel]

[crimson sedge]

What is this

[gentle lintel]

(quietly leaves with l hopital)

[fallen heath]

meanwhile me who hates l hopital but getting admit to hospital coming monday

[jaunty mural]

its not hard man

[fallen heath]

[jaunty mural]

you literally stick the series in

see most of the terms die

as x -> 0

in fact the only terms which won't die are things with a power of 3 or less

since u divide by x^3

[dire geode]

What kind of application are you looking for? It's used in all the sciences

[jaunty mural]

social science

[gentle lintel]

these

[jaunty mural]

ok genuinely, just try this

[fallen heath]

oh there's another I just recalled

[jaunty mural]

lookup the series for tan and sin

[gentle lintel]

(i think calc1 manipulation is enough for this)

[jaunty mural]

...........

just try it godamnit

[fallen heath]

$\lim_{x \to 0} \frac{\sqrt[7]{1+x^7} - \sqrt[3]{1 + x^3}}{x}$

[fallen heath]

or something like this( sry if this turns out unsolvable )

[jaunty mural]

very nice very nice

[dire geode]

What makes an application clever

[gentle lintel]

[jaunty mural]

you know

you can also integrate/differentiate termwise

of a taylor series

[fallen heath]

oh perfect! here you go another use for taylor series lmfao

[jaunty mural]

theres some theorem that says the radius of convergence is unaffected by integration/differentiation iirc?

Differentiate series for sin.. oh look here's cos

[jaunty mural]

Maybe this only applies to analytical functions or something like that (but almost everything you're interested in is analytical bar a few misbehaving points)

[fallen heath]

Find the coefficient of $x^{11}$ in the expansion of $(1 - x^{11})^2(1 - x)^{-1}$

[dire geode]

$\int e^{-x^2} dx$ for calculating values of the cumulative Gaussian distribution function

[wraith dagger]

riemann

[fallen heath]

:O

[jaunty mural]

Oh yeah so that reminds me

anything you can't integrate

to elementary functions

you can still find a taylor series for it.

[gentle lintel]

so series representation?

[jaunty mural]

exactly the application here

[fallen heath]

$e^{-x^2} = $

[gentle lintel]

not me being hell bent on finding a calc1 way to do this

[fallen heath]

,w \lim_{x \to 0} \frac{\sqrt[7]{1+x^7} - \sqrt[3]{1 + x^3}}{x}

[jaunty mural]

$$\int e^{-x^2}\dd x$$

[wraith dagger]

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

Requested by Ansh#3502

Shuri2060

[jaunty mural]

You find a calc1 way to get this

lel

[fallen heath]

,w limit [(1 - x^7)^{1/7} - (1 - x^3)^{1/3}] / x

[wraith dagger]

Results provided by WolframAlpha

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Requested by Ansh#3502

[fallen heath]

Lol

[jaunty mural]

,w integrate e^(-x^2)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Shuri2060#9525

[gentle lintel]

[gentle lintel]

erf

[jaunty mural]

indeed

we have to invent a function erf

but you could have figured out the series for it

[gaunt hamlet]

That's the sound someone makes when trying to integrate that calc 1 style

[fallen heath]

Ei(x) ?

[dire geode]

Practically all of asymptotic analysis is just Taylor series and perturbation

[fallen heath]

or maybe this: $\int \frac{e^x}{\ln x} \dd{x}$

[jaunty mural]

So suddenly we can integrate uh... practically anything

[gentle lintel]

(integration by parts?)

[jaunty mural]

unlike before

,w integrate e^x/ln x

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Shuri2060#9525

[jaunty mural]

^

[gaunt hamlet]

Who wants to name this one

[jaunty mural]

chrom(x)

because chrom will find the series for this

[dire geode]

Ansh(x)

[gentle lintel]

will not

[dire geode]

https://en.m.wikipedia.org/wiki/Asymptotic_expansion

[fallen heath]

coz riemann already has something after riemann's name

[dire geode]

@jaunty mural do you remember how the sech(z) laurant series formula is derived?

[jaunty mural]

never did that

[dire geode]

It looks like a difference of squares

[jaunty mural]

but i would imagine u derive it the usual way 👀

,w laurent series sech(z)

[wraith dagger]

Results provided by WolframAlpha

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Requested by Shuri2060#9525

[jaunty mural]

heheh

[fallen heath]

2! 4! 6!

[jaunty mural]

sinh(iz) = sin z ?

cosh(iz) = cos z

was it this?

maybe i wrote it wrong way round

sech = 1 / cosh = 1 / cos(ix)

[dire geode]

Hmmm I'll have to look in my complex analysis book

[jaunty mural]

u dont just do this do u

[fallen heath]

$\sum_{k = 1} (-1)^{k-1} \frac{(2^? - 3)z^{2k-2}}{(2k-2)!}$

[dire geode]

Doesn't residue theorem use Taylor series?

[gentle lintel]

(what math are you guys learning rn)

[jaunty mural]

💤

#groups-rings-fields

[fallen heath]

💤

[jaunty mural]

It came in my notes, so likely

laurent series has an annulus of convergence

which comes from 2 taylor series in a way

you sub x -> 1/x for the negative powers

[dire geode]

I guess integrals like $\int_{\gamma} \frac{dz}{z} = 2\pi i$ where $\gamma$ is a circle of radius $R$ containing the origin, are interesting

[jaunty mural]

summing 1/z over everywhere

🤔

?

or is that a normal C

[dire geode]

Complex analysis is great

[jaunty mural]

surely has to be C 😂 not \mbb C

[wraith dagger]

riemann

[jaunty mural]

👌

[cedar kiln]

@gentle lintel Has your question been resolved?

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[real bone]

I am trying to find dr/dt times 180 right?

[cedar kiln]

@real bone Has your question been resolved?

[shy furnace]

first, youre trying to find the ratio between volume and time in dependance on r and t

this is when the equation for V comes to help

because by dividing it by t, you get exactly that

after that, you set V/t = 4 try to solve for r

what you get as the result is the radius in dependence to t

after that, you just derivate it in respect to t

and to finally get the result, you put in t=180 in the derivative

[real bone]

dont I have to find the radius at t=0 which would be a volume of 4?

so 4=4/3(pi)r^3

[shy furnace]

not exactly

[real bone]

i mean t= 1

[shy furnace]

4 is V/t and not the volume itself

[real bone]

i just figure that the volume will be 4 after 1 second though right?

[shy furnace]

which is why you have 4=4/3πr³/t

[shy furnace]

yes

[real bone]

well you have to find r

so my calculations say r = (3/pi)^1/3

and with that you would then try to plug that into the v' equation

[shy furnace]

that is true

however

this would be the radius at t=1s

[real bone]

oh

and you cant just multiply by 180 can you

[shy furnace]

you cannot

[real bone]

ok

[shy furnace]

because if you do

you assume that it increases linearly

which it doesnt

[real bone]

its hard for me to think of this because t is not a variable

so will the answer still have an r in it because we do not know r at t=180?

[shy furnace]

we can know

because we know the formula for the volume

[real bone]

well it does say 4 per second

[shy furnace]

yes

[real bone]

and 180 seconds

[shy furnace]

if you divide the formula by t

[real bone]

is 720

[shy furnace]

you get the rate of change of the volume in dependance on time and radius

[real bone]

could you just do 720= volume formula to find r at t=180?

[shy furnace]

you could

but what you get as a result would be only the radius itself

and not its rate of change

as asked in the question

[real bone]

well i would take that radius measurement and plug that into the derivative of the volume forumal

formula*

i guess there are a couple ways to do this

[shy furnace]

but in respect to what?

[real bone]

time

my way didnt work

[shy furnace]

but you cannot isolate the volume without removing the time i think

[real bone]

so your way is finding r by dividing the volume formula by 180

[shy furnace]

my way would be dividing by t

without any value

plug in 4 for V/t

solve for r

find dr/dt

and plug in 180 in there

[real bone]

4=(4/3pi(r)^3)/180 ?

or i think i have to do 4= the derivative of v over t

[shy furnace]

i dont think so

[real bone]

oh wait

i know what you mean

[shy furnace]

because you wouldnt take the derivative of v

yeah

[real bone]

i mean the forula

but im gunna find r first

r= (3t/pi)^1/3 right?

[shy furnace]

yes

[real bone]

ok so now i take that and plug that into the derivative of the formula set equal to 4 right?

so 1/pi(r)^2 = dr/dt

[shy furnace]

not equal go 4

you already put in V/t before

what you have now is r/t

[real bone]

i got the right answer

[shy furnace]

oh nvmd

i misinderstood

[real bone]

no worries thanks for the help

[shy furnace]

no problem

[real bone]

.close

[cedar kiln]

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[knotty pasture]

need help

[knotty pasture]

How do I solve this summation

[soft cloak]

Do you know the formula for a geometric partial sum?

[knotty pasture]

I do not

[soft cloak]

i suggest you check that out beforehand then

because there really isn't much to this beyond that

unless you're tasked with finding the limit?

[knotty pasture]

I just want to represent this summation in algebraic normal form

[soft cloak]

what do you mean by algebraic normal form

[knotty pasture]

for instance like this,

[dire geode]

Apply this formula
https://www.media4math.com/sites/default/files/library_asset/images/Definition--SequencesSeries--GeometricSeries.png

[knotty pasture]

So would it be something like this?

[dire geode]

Looks good to me

[cedar kiln]

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[unkempt crest]

I often see that people call “lines drawn on” (subsets) of a space X “paths” while at the same time defining them as functions while the visualisation suggests that they are images if those functions and not the functions themselves. Is that an example of abuse of terminology?

[soft cloak]

i mean, in the same way that graphs aren't actually functions but images of functions

if you consider this abuse of terminology, then this "path" ordeal also is, but that's up to you

@unkempt crest

[cedar kiln]

@unkempt crest Has your question been resolved?

[unkempt crest]

Points on graphs aren’t of the form (x, f(x)) while elements of the image of the form f(x)?

[soft cloak]

Sure

But as far as path are concerned, to accurately plot a path would require 3D, which is quite cumbersome

[unkempt crest]

So what we look at is actually a projection of the path?

[soft cloak]

So, yes, only f(x) is shown to you, but that's partly because plotting x is both uneventful and annoying

[soft cloak]

No, i'd say it's the output of a function with 2D values being shown alone

without a graph of the input

(although, try to imagine how that'd even work for a second)

[cedar kiln]

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[jaunty mural]

plot it point by point.

[steady imp]

that is true

It is a horizontal translation to the left yes

[jaunty mural]

Why don't you try plotting point by point instead of asking

Surely you cannot be unsure if you try this.

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[tribal quiver]

So this is a bit more of an algorithm design question, but I'm trying to see if there's a combination of two coins (4 and 7) that will add up to another amount. The well-known dynamic programming algorithm runs in O(n) time, but I was thinking that we could maybe do a lot better by rephrasing the question into positive integer solutions for 4a + 7b = c, with c = 18 (solution is 7 * 2 + 4 * 1) and c = 17 (no solution), for the sake of discussion

[tribal quiver]

Looking at it, this also gave me the impression of a something-related-to-extended-Euclid-GCD problem, but with the added constraint that the coefficients need to be positive

I also found this: https://stackoverflow.com/a/27460171/4414853, but for some reason it doesn't seem to work when I did the calculations.

Is there some precondition for the suggested solution that I'm not aware of?

<@&286206848099549185>

Okay, I figured out what I was doing wrong. For posterity, I had forgotten to multiply my coefficients by c/gcd(a,b)

It works as expected.

.close

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[digital vortex]

the question is

1+3 = x find x

how do i do this question?

[foggy merlin]

1 = cos²(x) + sin²(x) = log(e)

[digital vortex]

ohhh

[dense wing]

ln(e)* smh

[foggy merlin]

yea if you want, but fellas here told me that the notation log is more used than ln

[digital vortex]

wait but couldnt you also flip the equation

like

1+3 becomes
3+1

[dense wing]

Yes, + is commutative.

[digital vortex]

yeah

so

if i were to do

3+1+1 = 5 right

then because i added an extra 1

[dense wing]

yes, congrats on passing kindergarten

[digital vortex]

5-1 = 4

so the answer is 4?

[dense wing]

Obviously.

[digital vortex]

ok but

couldnt you also

do it like this

3= 1*3

[dense wing]

sure

[digital vortex]

1= 1*1

so

[dense wing]

I don't give a crap about stupid stuff

[jaunty mural]

what is going on

[dense wing]

Lowlevel troll trolling

[foggy merlin]

just a troll

[digital vortex]

1 * 1 + 1 * 3

[dense wing]

[digital vortex]

and then factorise

1 * (1+3)

[foggy merlin]

they think they are funny but nah

[gaunt hamlet]

Sure, but you forgot to consider the Zeta function

[digital vortex]

.close

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[obsidian coral]

Dirac Delta function

[cedar kiln]

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[hardy umbra]

I need some guidance on where to start

[cedar kiln]

@hardy umbra Has your question been resolved?

[hardy umbra]

<@&286206848099549185>

[glossy halo]

I think you need to start out by finding out some stuff you can calculate: total volume the tank can hold, then find the volume currently in the tank (using similar triangle formulas ratios)

[hardy umbra]

With the similar triangle formulas ratio, do I not need the diameter of the water surface to find the smaller "cone"? Or what do I relate between the two? I think im blanking somewhere.

[glossy halo]

you have the heights of each, isn't it just the D/X = (smaller d)/w ?

[hardy umbra]

Youre totally right.

[cedar kiln]

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[hardy umbra]

.close

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[eternal patrol]

which of the following points would fall on the line produced by the point-slope form equation y-12=2(x-6) when graphed

A (1,-2)
B (2,-4)
C (0,1)
D (-2,-4)

[glad kestrel]

looks awfully like a quiz question now that i think about it

[eternal patrol]

nah its math hw

[glad kestrel]

can you take a picture of said homework

[eternal patrol]

im on pc

so no

[glad kestrel]

multiple choice math homework

[crimson sedge]

use a snipping tool

[glad kestrel]

well if someone else wants to help you they can do as they please but seems too sus to me tbh

[eternal patrol]

then dont answer

.close

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[crimson sedge]

😔

[cedar kiln]

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[jaunty mural]

You couldve probably factorised to make your life easier

but its not nice algebra either way

well that 2 on the ab makes it not nice to factor either

not much you can do.

you can get something like wolfram to check for you idk

[dense wing]

Refer to what Shuri said

Use wolfram or any other CAS to check your algebra work

[cedar kiln]

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[crimson sedge]

Hey

[crimson sedge]

My work so far

Don’t really know what the next step should be

[gaunt hamlet]

What is your understanding of the method of lagrange multipliers?

[crimson sedge]

I know the largest value yielded the constrained max and the smallest is the class strained minimum

[gaunt hamlet]

Now, I have no formal education on the subject, but my understanding of lagrange multipliers is that if you have a function f constrained by g, then the extreme values are given by ∇f = λ∇g

[crimson sedge]

What does that mean

[gaunt hamlet]

The gradient of f is some constant multiple of the gradient of g at the extreme values

[crimson sedge]

The way we learned in class has us using systems of equations

Is that the same method

Or a different one?

[gaunt hamlet]

It will become a system of equations, yes

[crimson sedge]

I know the process I’ve taken is the way we’ve been doing it in class

Is ur way different?

[gaunt hamlet]

I'm not familiar with your process. I only know the process given in my textbook

They might be the same, but I can't tell. In my textbook, we'd end up with the system of equations ∂f/∂x = λ ∂g/∂x, ∂f/∂y = λ ∂g/∂y, and g(x, y) = 0

Three equations, three unknowns

[crimson sedge]

Ok then what

[gaunt hamlet]

Well, I've never heard of this method, but it's probably best to use it. I'm sure both methods are basically the same thing written differently anyways

So you have three equations, F_x, F_y, and F_λ, and all three are 0

[crimson sedge]

Yeah that’s where I’m at rn

[gaunt hamlet]

Thought about it for a bit, both methods are exactly the same

[crimson sedge]

Got all three partial derivatives

[gaunt hamlet]

You have three equations and three unknowns. Do you know how to solve a system of three equations?

[crimson sedge]

Nope

[gaunt hamlet]

Well, just like with 2 equations, there's a variety of methods

There's a form of elimination, and then there's substitution

Which would you prefer?

The elimination method would end with us eliminating a variable from two equations, leaving a system of two equations

The substitution method would end with us solving for one of the variables on one of the equations, and plugging it in to the other two, again giving a system of two equations

[crimson sedge]

Which one would be the easiest

[gaunt hamlet]

Well, the elimination technique is used pretty much exclusively in linear algebra.

So we could do that

You know how in elimination with two equations, we add or subtract the two equations to eliminate a variable?

It's the same principle

[crimson sedge]

Yeah sure

So multiply the second equation by 2

And the x will cancel out right

[gaunt hamlet]

After adding the first and second, yes

[crimson sedge]

Wait what?

Wait let me write it down

[gaunt hamlet]

Multiplying by 2 doesn't just remove a variable

It's like in elimination with 2 equations. To eliminate x in 2x + y = 0 and -2x - 2y = 2, we have to add the two equations to get -y = 2

[crimson sedge]

3y +5 lambda =0

[gaunt hamlet]

You know what might be easier. Instead of cancelling out the x, notice the 3rd equation doesn't have a lambda

[crimson sedge]

Yeah

Cancel out the lambda?

[gaunt hamlet]

Yes

Then we have a system of two equations in x and y

The third equation will always lack a λ, so it seems cancelling a λ will always be the best option

[gaunt hamlet]

,rotate

[crimson sedge]

Sorry I don’t know why discord sends my stuff sideways

[gaunt hamlet]

Positive 5x?

[crimson sedge]

Negative sorry

[gaunt hamlet]

Then yeah

[crimson sedge]

Solve for y

[gaunt hamlet]

And you have the third equation which just contains x and y, and you can perform your normal methods for system of equations

[crimson sedge]

X = 4/5 y

Plug in 4/5 y for x?

[gaunt hamlet]

Yep

[gaunt hamlet]

-7/5?