help-17

[patent nymph]

what is f(2)?

[tidal trail]

2

g(2) is also 2

[patent nymph]

you can use the product rule here

[tidal trail]

4?

[patent nymph]

How did you get that?

[tidal trail]

Straight up 2 times 2

[patent nymph]

try using the product rule for derivatives

,tex .diff rules

[twin meteor]

even order group => solvable

[tidal trail]

8?

4+4

[patent nymph]

yes

[tidal trail]

What abt this

[patent nymph]

you can use the differentiation rules

,tex .diff rules

[twin meteor]

even order group => solvable

[tidal trail]

?

[patent nymph]

you can start by using the chain rule

[tidal trail]

I got sth like this

Wait that's the same thing...

[obtuse sierra]

show your work

[tidal trail]

I deleted it...

Lemme try again

[vocal sleet]

@tidal trail Has your question been resolved?

[tidal trail]

Nvm I'll ask again ltr... at the doc's

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[vocal sleet]

@opal ridge Has your question been resolved?

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@opal ridge Has your question been resolved?

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[elfin moon]

b^2-4ac >=0

[obtuse sierra]

show what you did

its 1/81

[elfin moon]

i did this

16+4 log a base 3>=0

[paper depot]

log_3(a)

[harsh canopy]

why the x^2?

[paper depot]

also yeah, b isn't -4x, it is just -4.

[elfin moon]

hmm edited

so i have to give value to log

to make it real

[harsh canopy]

well if you solve that inequality it should give you the correct answer

[paper depot]

no you're overthinking it.

you have an inequality

$16 + 4 \log_3(a) \geq 0$

solve it

[twin meteor]

Ann

[elfin moon]

got it

1/81

a=1/81

[paper depot]

what's this

[elfin moon]

yes

3^(-4)

which is 1/81

[paper depot]

"what's this?" is NOT a yes/no question.

also that doesn't look at all like 3^(-4) to me.

[paper depot]

the (-4) is too big and too far above the 3 and not enough to the right.

[harsh canopy]

get out of here

theres no way you use white mode

[final zephyr]

sorry

[harsh canopy]

how

[final zephyr]

your eyeballs shall BURN

[harsh canopy]

😭

[paper depot]

@final zephyr @harsh canopy please take this chatter to #discussion.

[elfin moon]

this is my rough book

[paper depot]

1. you are still presenting it to others, so it has to be presentable and readable. otherwise, you are saying "I want you to read this work. I have made it hard to read on purpose. Suffer!"

[paper depot]

2. even in your rough work you should not commit such notational idiocy.

unless you want your future self to suffer as well.

and the solution to the inequality is a ≥ 3^(-4). though yes, 3^(-4) itself is the least value of a.

[elfin moon]

I don't think anyone will get confusion with this

[harsh canopy]

bro is getting cooked for writing something 2mm too high

[elfin moon]

I have read enogh 10* times bad handwriting of french/europeans work than me

[paper depot]

it is not 2 mm.

[paper depot]

i was confused.

[harsh canopy]

ok 4mm at best

[final zephyr]

ann deep-fries people for not knowing maths it is quite predictable behaviour from her

[elfin moon]

i want to fly (-4) in the sky with kite🪁

i write things 10000 times clear and bigger in size compared to europeans

[harsh canopy]

suffer

[paper depot]

@harsh canopy please take this banter to #discussion.

[paper depot]

size was never the problem

the problem was placement

[elfin moon]

this is log not LOY

[elfin moon]

I write like this in my books

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[elfin moon]

What to do for the 4th question?

<@&286206848099549185>

[vocal sleet]

@elfin moon Has your question been resolved?

[elfin moon]

Why are they not including 0 in range?

[obtuse sierra]

well its a reciprocal function

cant be zero

,w plot 1/x

[twin meteor]

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[elfin moon]

yes true

But u didn't understand my actual doubt

The function is not 1/x here

They should include that value which makes it zero

[worthy citrus]

What x value would make the function = 0? If you think 0 should be in the range

[elfin moon]

I meant how did they check this interval (-4/5 to 0)

[elfin moon]

u tell me how you will choose these intervals without help of desmos/wolfram

@obtuse sierra u too

we do not have much time here to check all the random points if this is your view of chekcing intervals

[obtuse sierra]

checking the end points of the range of the function in the denominator

it ranges from [-5/4,infinity)

plug in -5/4

gives -4/5

this will be one end point of an interval

now plug in 0 but from left side

[elfin moon]

this is clear for me

[elfin moon]

hmm

[obtuse sierra]

gives -infinity

im not saying x=0 but the whole function in the denominator

[elfin moon]

where is it giving infinty at 0

it gives -1

[obtuse sierra]

this

[elfin moon]

1/4-5/4=-1

[obtuse sierra]

,w plot (x-1/2)^2 - 5/4

[twin meteor]

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[obtuse sierra]

if we plug in whatever the positive root is from lhs

it will give us 0 but from negative side

you know limits right?

[elfin moon]

yes i know it

[obtuse sierra]

you understood what i said?

[elfin moon]

no completely but keep typing

[obtuse sierra]

ok, so one interval i got is (-infinity,-4/5]

[elfin moon]

-5/4? or -4/5

[obtuse sierra]

sry

now when the function tends to infinity

the fraction tends to 0

it isnt 0 but tends to

[elfin moon]

true

[obtuse sierra]

similarily if i again plug in the positive root from rhs, it will give me 0 from positive side, so fraction will tend to positive infinity

so next interval becomes (0,infinity)

[elfin moon]

what about (-4/5 to 0)

how did you discard it?

[obtuse sierra]

i didnt

ah idk how to say this

i just formed the intervals, and those values werent in them

[elfin moon]

did you read this?

[obtuse sierra]

if you still wanna crosscheck, equate the function to any value in that interval

[obtuse sierra]

i explained every process which i use

i never use graph plotters

dont have that much time

[elfin moon]

true

i will read it agagin and again

[elfin moon]

I didn't understand why they wrote (e^x+1)/(e^x+x) to e^x/e^x+1

<@&286206848099549185>

[worthy citrus]

Lhopital

[elfin moon]

Ohh

And again they did so they got

[worthy citrus]

Big giveaway being the big H above the equals sign

[elfin moon]

E^x/e^x

[elfin moon]

Which means?

[worthy citrus]

Hopital..

[elfin moon]

Any other method?

[elfin moon]

To solve this question?

[worthy citrus]

Guess

[elfin moon]

Expansion of e^x?

[obtuse sierra]

wont help much

wont cancel anything

also its all inside ln

[elfin moon]

I got binomial method

Limit tends to infinity

[obtuse sierra]

yeah this is correct

but with a condition that |x/e^x| < 1

[elfin moon]

What if x tends to 0

[obtuse sierra]

it should be e^2

its 1^infinity form

[elfin moon]

i did not get it

we took common

[obtuse sierra]

yeah got it

that are just the first two terms of the expansion

binomial expansion has infinite terms

when you will write some more terms

and sub in 0

you will see a pattern, which will form the expansion of e^1

so final answer will be e*e = e^2

[elfin moon]

Ohh got it

1+x =e^x
e^1 =1+1

[obtuse sierra]

no

e^1 = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + .....

[elfin moon]

Yes

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[fiery zenith]

Hello is 0.005% 1/20000?

[mild flower]

    1% = 1/100
  0.1% = 1/1000
 0.01% = 1/10,000
0.005% = 1/20,000

[fiery zenith]

thank you so much

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[fast quarry]

where have i gone wrong here

[fast quarry]

thats the answer

[split wind]

-2x⁵ instead of -2x⁶

[fast quarry]

oh yes

i see now

silly mistake

thanks

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[minor blade]

i cant find the answer and im probably not wording it properly, but if i have x! and x^x for x->+infinity, which one is of the highest order of infinity?

[split wind]

i think it's hard to answer your question without know what is "order of infinity"
https://www.quora.com/Are-there-many-orders-of-infinity

but in case you are just wondering which will be bigger as x tends to +infty,

just check
lim x to infty, x^x/x!

[minor blade]

ah you were right, i completely forgot i couldve jsut checked like that ^^" tysm, also the link seems interesting, thank you!

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[languid helm]

I am working through zill's precalc with calculus previews and want to confirm if my solution for one of the problems is correct.
The question is in attached screenshot

[languid helm]

my solution:

One of the points in the maximum vertical distance is the vertex of the parabola, so we start with finding that

h = -b / 2a = 3

Now, put 3 in equation of parabola, we get the vertex (3,9)
Now since line is vertical, we put x=3 in equation of line, and get y=3

so max distance = distance between (3,9) and (3,3) = 6

but the answer mentioned here is 72 - https://quizlet.com/explanations/textbook-solutions/precalculus-with-calculus-previews-6th-edition-9781284077308/chapter-2-exercises-52-07ea278a-1bf4-40c4-8d4b-60f93693c623

Looks like they have messed up with substitution of values in the parabola equation

Please let me know about this

Thanks.

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[digital eagle]

Why is "a" being treated as a constant?

[hidden kelp]

because it is. The limit is in terms of x, so any other variables are constants

[digital eagle]

so when I take the limit in terms of x and differentiate I treat any other variable as a constant and don't use the product rule for example?

for other limits

as an example

[hidden kelp]

exactly! Think about it, if I asked you to do: $\dv{x}(3\cdot x)$. You would bring the 3 to the outside because it is a constant, right? Well it's the same thing if I asked you to do $\pdv{x}(t\cdot x)$

[twin meteor]

imTyp0

[digital eagle]

yea t

[digital eagle]

thanks :)

[hidden kelp]

it all depends what variable you do your operation on. The rest is just considered constants :)

[digital eagle]

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[elfin moon]

true?

[worthy citrus]

well you can see it yourself from the graph

[elfin moon]

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[wary aurora]

Hey I managed to factor 24a^2 +14ab -20b^2 into (8x+10y)(3x-2y) but only by extremely tedious trial and error testing different values for the coefficients. Is there a mor intuitive or systematic way to factor this expression?

[whole oasis]

yes it is

[wary aurora]

Whoops, answer should be (8a+10b)(3a-2b)

[whole oasis]

we can treat it as quadratic

with one parameter

which is b

[wary aurora]

Im not sure what you mean when you say parameter

[whole oasis]

treat b as constant, some number

and a as variable, like we often treat x

[wary aurora]

oh, then multiply it out and use the quadratic formula?

[twin meteor]

Modus

[whole oasis]

and now

you can find the roots

and use a(x-x1)(x-x2)

[boreal remnant]

ac method

factors of -20*24b^2, add to equal 14b

you can even ignore the b's here. factors of -480, add to equal 14

[wary aurora]

oh sorry, my phone restarted by itself

Hey @whole oasis when you say use a(x-x1)(x-x2) whqt do you mean? I havent aeen that form before, im not aure which value go where

[whole oasis]

have you calculated roots

[wary aurora]

You mean like quadratic roots or?

[whole oasis]

yes, roots of the quadratic

[wary aurora]

I know how to use the quadratic formula if thats what you mean

[whole oasis]

yes

we can use the quadratic formula

[wary aurora]

Where 24 is a, 14 is b, and -20 is c?

i get 2/3 and -5/4 as roots

[whole oasis]

don't forget about b

"a" is 24
"b" is 14b
"c" is -20b^2

(as bolded above)

[wary aurora]

Ah

[whole oasis]

roots are supposed to be in terms of b

[wary aurora]

One sec, let me try the new values

I end up with -14 +- sqrt(-1724b^2) /48 but 1724 isnt a perfect square

[whole oasis]

it should be

$$\frac{-14b \pm \sqrt{2116b^2}}{48}$$

[twin meteor]

Modus

[wary aurora]

when i initially pulg the values in for the sqrt i have (-14b)^2 - 4(24)(-20b^2)

thats the b^2 - 4ac part

[whole oasis]

,w (-14b)^2 - 4(24)(-20b^2)

[twin meteor]

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[wary aurora]

Oh i got my signs mixed up

1920 ended up being negative for me, my bad

So the roots are now -14 +- 46b / 48 right?

[whole oasis]

-14b +- 46b / 48

so

[wary aurora]

Oh right

[whole oasis]

as you got them here

but with "b"

2/3b and -5/4b

[wary aurora]

So -60b/48 and +32b/48 which then simplify to those

Okay, so what is the next wtep that gives us the four coefficients from those two roots?

[whole oasis]

you know factored form of a quadratic?

[wary aurora]

two binomial factors?

[whole oasis]

that stuff

in other words

leading coefficient multiplied by (x - 1st root) * (x - 2nd root)

this method might seem hard, but it's faster than trial and error, I understand it is new for you

[wary aurora]

What is the value of a for this case?

[whole oasis]

the same as it was in the quadratic formula

= 24

[wary aurora]

oh

And x corresponds to?

[whole oasis]

"x" is variable, in our case "a" is variable

but just "a", not 24 (don't confuse them, just bad notation in this case)

[wary aurora]

Ok, so it would be 24(b- 2/3b)(b+5/4b) ?

[whole oasis]

24(a- 2/3b)(a+5/4b)

since "a" is variable, sorry for my edits

now you can multiply the brackets by 3 and 4, to get rid of the fractions

and you'll get your factorization

[wary aurora]

Wait, how can we multiply 3 by (a- 2/3 b) without it affecting the other terms?

[whole oasis]

multiplication is commutative

you can multiply any two terms you want

[wary aurora]

Oh, so break the 24 into (2)(3)(4) ?

[twin meteor]

Modus

[wary aurora]

Ah, okay

Wow, its so crazy that this problem is from the first set on page 5 of my algebra book

[whole oasis]

well, this method might be new for you, this is why it seems hard and takes a long time

normally it goes quickly

maybe a few more examples and you'll see on your own

[wary aurora]

Okay, the factored form of a quafratic is completely new to me. Ive never seen it before in my classes. Oh, and thanks for your help btw, i appreciate it alot

[whole oasis]

if you don't mind

we can also solve this using other method

I'd say faster, but more tricky

[wary aurora]

Oh? Sure, id like to see it

[whole oasis]

so

it reminds me of the formula

a^2 + 2ab + b^2

= (a+b)^2

[wary aurora]

Ya, i saw that too. But i couldn't intuit the coefficients

[whole oasis]

notice that

we can rewrite it as

24a^2 + 14ab + 49b^2 - 69b^2

why exactly that?

the middle term is a key

(let's use x^2 + 2xy + y^2 = (x+y)^2 not to confuse the letters)

[wary aurora]

Okay, but how do we infer what the coefficients would be? Is there any binomial that when squared spits out 24a^2 + 14ab +49b^2 ?

[whole oasis]

we want to have a pefrect square

from the middle term

[wary aurora]

But because it expands to (a+b)(a+b) the coefficient of a would have to be the square root of 24,.which isnt an integer

[whole oasis]

okay, in this case it leads to fractions

I guess it wasn't a good idea

[wary aurora]

Ye, is weird problem

[whole oasis]

hehe

but it's fine

[wary aurora]

If i was handed this book in 7th grade i would have died lol

If its too much of an issue no need to worry about it, i can close the channel if you want

.close

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[verbal merlin]

i have solved part a

now for part b i wrote out this lemme take a pic

so i have dis right now for part (b)

,rotate

[verbal merlin]

im not sure if im doing it the right way, plus how wld i know what t is o.0?

if thats not neat i can type out in symbolab if it helps ^w^

[vapid summit]

looks to me like the last term of f(t) isn't used. So what values of t will make that sin go to 0?

[verbal merlin]

oh so u mean the -pie(sin(4t))? rightt

so i have to find what value of t will make the sin4t go 0?

[vapid summit]

I think so

[verbal merlin]

t = π

is the 4cos4t used?

[vapid summit]

I actually have to go, I'll come back in like 15 mins

[verbal merlin]

alright sure

if anyone else here avail to help please do if not ill wait for Inky ^^

[queen ermine]

You chose t= pi?

[verbal merlin]

is it wrong o.0

[queen ermine]

No i think it’ll work 🙂

[verbal merlin]

o.0 really?

im so sus of math my apologies TvT

[queen ermine]

🙂

What are the values of sine and cosine at integer multiples of pi?

[verbal merlin]

wait i dun get what u mean by this im sorry

Wait also how exactly do i write this and solve dis oml

i am quite new to this topic

[queen ermine]

This all from the unit circle

[verbal merlin]

oh

wait er im a bit lost

what do i do now exactly?

[queen ermine]

You need a formula for sin(nPi) and cos(nPi) when n is an integer

Compute some examples

Sin 0, sin pi, sin 2pi, etc are all 0

What about cos 0, cos pi, cos 2pi, cos 3pi?

There’s a pattern that you can write down in a simple formula in terms of n

[verbal merlin]

oh

cos 0 = 1

cos pi = -1

2pi is 1

3 pi is -1

so its alternating between 1 & -1

how do i write the pattern tho @queen ermine

<@&286206848099549185>

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[crimson pumice]

can someone help me understand the logic of this step?

[crimson pumice]

the answer is the 2nd image, the context is the first image

[paper depot]

^

[crimson pumice]

oh

[mystic skiff]

[crimson pumice]

but how does that turn into siny?

[mystic skiff]

that's just a property of sine

[paper depot]

gamma not y

also

two things

[crimson pumice]

bro ann I know I'm not typing up the symbol for gamma if y can portray the same idea

[paper depot]

1. sin(π - x) = sin(x), as can be verified by several means
2. why did you lie about which step you were confused at? you asked about one step and then immediately asked about another.

[paper depot]

please don't call me bro.

also just typing "gamma" is not a finger-breaker.

[crimson pumice]

but y and γ are so similar

[paper depot]

the resemblance is superficial.

[crimson pumice]

well alright then guess my question is answered

can't deny that

[mystic skiff]

you can prove that sin(pi-x) = sinx with the sine subtraction formula

[paper depot]

or by looking at the unit circle and noting what happens when you flip an angle across the y-axis

[crimson pumice]

ah I see, that makes sense

well thanks for the help

.close

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[river kettle]

In log differentiation why is everything raised to e?

[river kettle]

Or I mean why does e become the base

[boreal remnant]

ln and exp are inverses of each other

so exp(ln(something)) = something

[river kettle]

you mean e is inverse of ln?

[boreal remnant]

ye

we use exp() a lot too in notation to refer to e^

[river kettle]

oh I’ve never seen that one before

[paper depot]

bad wording

the functions e^x and ln(x) are inverses to each other.

[paper depot]

also it's not "everything raised to e" but "e raised to everything"

the reason e^x is so often used in this way for differentiation is that e^x is its own derivative. so applying the chain rule is very easy.

[river kettle]

I corrected myself stwr

After

Where does x^2•ln2 come from?

[river kettle]

Why is that brought down from the power

[regal slate]

chain rule

[river kettle]

ok I see

[river kettle]

What’s happening in the last 2 lines? I don’t understand

[river kettle]

From ere

[toxic crater]

remember you can write
(x^2)ln(2) as ln(2^(x^2)) by properties of logarithms

and e^ln(y)=y

that should help u understand the last two lines

in other words, $$e^{x^{2}\ln{2}}=e^{\ln{2^{x^{2}}}}=2^{x^{2}}$$

[twin meteor]

musikalischer mathematiker

[river kettle]

This is confusing

[paper depot]

they rewrote $2^{x^2}$ as $e^{x^2 \ln(2)}$ in the beginning, do you have any objections to that?

[twin meteor]

Ann

[paper depot]

@river kettle

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@river kettle Has your question been resolved?

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[vast shale]

Hi I need someone to explain something

[vast shale]

So I have a question

[dreamy viper]

then ask it

[vast shale]

That was like log of x to base e

Wait let me find it

[spiral turtle]

What about ln(x)?

[dreamy viper]

$log_ex=\ln{x}$

[vast shale]

Yeah that

[twin meteor]

Arctic

[vast shale]

Yes

The thing is in the example they said ‘put both sides to the power of e’

I don’t get how that works

[paper depot]

from $\ln(3x-1)=2$ go to $e^{\ln(3x-1)} = e^2$

[twin meteor]

Ann

[vast shale]

What

[thin vale]

if f(x1)=f(x2) then x1=x2

for f(x)=e^x

no two different x-values, produce the same output

[spiral turtle]

Sorry, too advanced (injective that is)

[vast shale]

Yes but how can they just put the powers

[thin vale]

think about this

let the left hand side = x1

let the right hand side = x2

since x1=x2

f(x1)=f(x2)

for f(x)=e^x

so you can raise both sides as a power of e

while garunteeing they are both still equal

if

5=5

e^5=e^5

[vast shale]

And cancelling the In?

[spiral turtle]

Essentially, as long as you do the same thing to both sides of an equation you preserve the equality. Provided that the thing you do isn't undoing something that can map two inputs to the same output.

[thin vale]

ln(x) is a log with base e

so e^ln(x)=x

because recall what ln(x) is actually doing

it is asking you, e raised to what power, equals x

[vast shale]

Yeah

But they cancelled it by putting both sides to a power of e

[thin vale]

mhm, because when you raise ln(x) as a power of e, you get e^ln(x) = x

[spiral turtle]

So if we have x = y, would you agree that if there is a function f, that f(x) = f(y)?

[vast shale]

It’s equal to x?

[vast shale]

Yeah

[spiral turtle]

Ok, now let's say f(x) in this particular instance is e^x

So if x = y then e^x = e^y

[vast shale]

Yeah

Wait so the f(x) isn’t e^x?

[spiral turtle]

So that's all they're doing here, and then some log based simplifications that Austin was working you through

[spiral turtle]

It is in this instance.

[vast shale]

Ok

[spiral turtle]

But in general f could be anything

Or anything that is defined at x

[vast shale]

So log of base a put to the power of a like e^In(x)=x is a logarithm rule?

[spiral turtle]

It's the definition of the logarithmic in fact.

Or at least, one of the definitions, and the most common elementary one

[vast shale]

What’s that

[spiral turtle]

e^y = x, we want to solve for y, but we can't. So log is defined as the function that inverses this, y = ln(x)

We raise both sides by e to get e^y = e^(ln(x))

And note that e^y = x as previously established

So x = e^(ln(x))

[vast shale]

Uh ok

[spiral turtle]

You seem lost or unconvinced

[vast shale]

A bit lost but I think I get it

Wait log x divided by log y where x=2y is 2 right

[spiral turtle]

log(x)/log(y) = log(2y) / log(y)
= (log(2) + log(y)) / log(y)
= 1 + log(2)/log(y)
= 1 + log_y(2)

[vast shale]

Oh

Ok

And log of x to base b convert to base 10 is log x/log b?

[spiral turtle]

https://media.discordapp.net/attachments/486844829209198613/1136895787696140318/246857845285453824.png

[vast shale]

What’s c in base change

[still mica]

The base you are converting to

[vast shale]

Oh ok

Thx

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[pallid jetty]

This is my attempt to do the question but Im not sure why I’m wrong. I tried sketching the graph and it still doesn’t seem to be an odd function. I dun understand how im getting ans of an odd function.

[vocal sleet]

@pallid jetty Has your question been resolved?

[pallid jetty]

<@&286206848099549185>

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[pallid jetty]

.reopen

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✅

[pallid jetty]

<@&286206848099549185>

[somber yew]

Not completely sure, but your omega seems to be wrong if compared with the answer key

Should be pi/L and not 2pi/L probably

[dull maple]

In general, omega is 2Pi/(time period) so what about that?

[somber yew]

Yeah I guess but for fourier cosine/sine stuff we only look at half the interval

Since the other half is the odd or even part

[dull maple]

Ah. Okay. That makes sense now.

[verbal merlin]

hello again

@dull maple can explain?

we dont really understand TvT

[pallid jetty]

when talking abt half the interval, I thought the omega value wouldn’t change?

[somber yew]

In your question L=2 which should be your fundamental frequency

so omega_0 = pi/2

[verbal merlin]

its not 2pi cus the thing is like halves?

[somber yew]

yes you are only given half of the interval

the other half (defined for -2 <= x < 0) is either the odd or even complement

but that one is defined implicitly when asking for the fourier sine or cosine series

[verbal merlin]

wait is it using dis?

[somber yew]

yes thats another formulation

but here T = 4

[verbal merlin]

wait

im a bit confused abt the pi/L thing

how does it come about?

[somber yew]

when they ask for a fourier sine series of a function over an interval it should be implicitly understood that it has been periodically extended and also made odd if only defined over an interval between 0 and L

its the same as a general fourier series where they extend the function over a period periodically

[pallid jetty]

ohh so in this case L is 2?

[somber yew]

uhh the time period is 4

you need to obtain pi/2 for the fundamental frequency

[verbal merlin]

Wait isnt t greater than 0 lesser than 2?

[somber yew]

yes but it has been extended

[verbal merlin]

Oh

[somber yew]

think about this way

sine is an odd function

[verbal merlin]

Mhm

[somber yew]

so a sum of sines

is also an odd function

[verbal merlin]

Mhmm

[somber yew]

so your function you are approximating better be odd or you'd have no chance

therefore if your function is defined over 0<t<2, you implicitly also define the function for -2<t<0 so that it is odd

[verbal merlin]

Ahh

So u get the 4 cus of that

[somber yew]

yes

[verbal merlin]

So issit like its 2pi/4 which reduced becomes pi/2?

[somber yew]

yes thats one way

[somber yew]

if you take T=4 and use these formulas you should arrive at the answer

see for example how T/2 is in the integral bounds

[pallid jetty]

oh so omega would be pi/2?

[somber yew]

yes thats what it should be

it's also that way in the answer

[pallid jetty]

I was wondering where did the 1-cos(n*pi) come from?

from the ans key

[somber yew]

it'll probably come out that way when you work out the integral with the right omega0

for the fourier cosine series you are basically calculating the series for this function:

and this function for the fourier sine series:

[pallid jetty]

correct me if im wrong. So for this qn, we have to form a graph such that we get both odd and even function. Afterwards we will be able to find A0 and An with our even function and Bn with our odd function

[somber yew]

yes thats maybe an easier way to go about it

first define its odd or even extension and then find the series for that one

[pallid jetty]

both range are from -2<x<2?

[somber yew]

yes thats one period

so T = 4

[pallid jetty]

alright I’ll work it out and let uk when I’m done

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@pallid jetty Has your question been resolved?

[dull maple]

Sorry. Was reading up on this myself.

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[stuck current]

Is there anyone who can help me with a problem in graph theory?

[vast shale]

Yep

[stuck current]

It's mostly a drawing problem, I spend 3 hours like an idiot trying to draw K8 into a two holed torus. Could you help me with a complete drawing of it, without relying on glueing sides.

This was the most I managed to do

[vast shale]

Introduce

[stuck current]

myself or what?

[stuck current]

What do you mean by introduce? Simply saying @vast shale ? It's simply a drawing problem that I could not do since I couldn't connect all vertices together without overlapping of edges

[vocal sleet]

@stuck current Has your question been resolved?

[vocal sleet]

@stuck current Has your question been resolved?

[vocal sleet]

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[fast quarry]

i dont really get this end state ment

i agree that whichever order u select the functions for integration by parts just repeats

but then i dont knowhow they have solved it

[hasty pulsar]

Bring I to other side

U get 2I=.....

Then divide by two

[fast quarry]

i dont really understand do you mind showing me the workings?

if you bring I to the other side doesnt it just = 0

[mild flower]

you have $I = e^x\cos x + e^x\sin x - I$

[twin meteor]

Hayley

[mild flower]

so if you add $I$ to both sides you get $2I$ on the left

[twin meteor]

Hayley

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[urban obsidian]

not sure how to do this tbh

_ _ _ _ _ _

[merry python]

what have you tried?

[urban obsidian]

to get 7.5 i would need a 7 and an 8

_ _ _ 7 8 _ _

so that works?

then the last two digits have to be 9 and 10

[merry python]

First of all write all the positive integers possible

[urban obsidian]

1-10 (exclusive)

[merry python]

Okay

and all numbers are distinct

so what do you think would be the smallest set of numbers

The smallest possible set would be 1,2,3,4,5,6

and what do think it's median is?

[urban obsidian]

3.5

[merry python]

So, did you come to a conclusion?

[urban obsidian]

yeah i did

but how would i think of that

like hmm

wait

to get an average of 15

i need a sum that yields 15

(6,9) and (7.8) is the possible way to get that

[merry python]

average of 15?

[urban obsidian]

both don't work

average of 7.5

my bad

i mean median 7.5

bro

okay

i conflated 10 different stuffs

whoops

[merry python]

Okay so wouldn't a sequence satisfying that be 5,6,7,8,9,10

[urban obsidian]

yeah

10 is not allowed though

lmao

[merry python]

10 is

[urban obsidian]

oh bruh

okay that solves it then

[merry python]

LESS THAN OR EQUAL TO

[urban obsidian]

i literally was trying to do it without 10

bruh

[urban obsidian]

even wrote exclusive here lol

damn

yeah

i edited it right after lol

anyway

i can try numbers or just consider lowest medians right?

both would work

_ _ 24 _ _ to get a median of 3

which means the first digits have to be a 0 and 1

0 is not positive so boom

not possible

[merry python]

Yes

[urban obsidian]

how would i do this in a quicker way?

10C - 2W = ? provided that C + W = 20

i just counted downards lmao

[sage merlin]

also C and W are positive

so just

[urban obsidian]

C W
20 0
19 1
18 2
17 3
16 4
15 5
14 6
.
.
.

i found 152

i found 140

didn't find 124

[sage merlin]

jeez

uh

since it’s multiple choice

just plug them in

[urban obsidian]

found 104 and found 80 lol

[sage merlin]

and see if they work

💀

[urban obsidian]

how?

there's two variables lmao

[sage merlin]

10C - 2W = [answer]

[urban obsidian]

yeah

[sage merlin]

plug in the multiple choices

for [answer]

alternatively

[urban obsidian]

isn't that exactly what i'm doing lmao

except it's much more harder i guess

[sage merlin]

12C = [answer] + 40

so [answer] must be 8 mod 12

which 124 is not

but everything else is

[urban obsidian]

wait lemme try that out

[urban obsidian]

could you elaborate what happens after this

where did you get that answer must be 8 mod 12

[sage merlin]

cuz

we know C is a positive integer

thus

12C is a multiple of 12

[urban obsidian]

yes

[sage merlin]

[answer] = 12C - 40

[urban obsidian]

and ans + 40 has to be a multiple of 12

[sage merlin]

yea

40 is 4 mod 12

12C - 40 is -4 or 8 mod 12

[answer] must be 8 mod 12

in order for it to be a possible score

I love number theory bro

[urban obsidian]

bruh

i haven't done number theory formally

but

-4 mod 12 is 8 mod 12 right?

40 is 4 mod 12 yeah

12C - 40 yeah idk what to do here

anyway i guess just answer + 40 has to be a multiple of 40 works equally well lol

nice

[sage merlin]

uh

[sage merlin]

me neither bro 💀

[urban obsidian]

bruh

anyway i get it lmao

12C is 0 (mod 12)

[sage merlin]

yea

[urban obsidian]

then -40 is just -40(mod 12) which is -4(mod 12) which is 8(mod 12)

💀 ain't doing alla dat doe 😭 just tryna be a simple dude and get da answer

so ima just do answer + 40 = multiple of 12

[sage merlin]

it’s pretty fast once ur used to it

yea same meaning

do what works best for u

[urban obsidian]

indeed

also small numbers

so i guess it doesn't matter

thanks btw

How would i do this? I guess this is a skewed triangle

is there a straightforward way to always get the altitude of any triangle?

[merry python]

You need to use heron's formula

do you know what heron's formula is?

[urban obsidian]

oh okay i see

use heron's formula to find area

then find height?

[merry python]

Yes

[urban obsidian]

but how do we know when a triangle is skewed or not?

Sometimes perpendicular bisector separtes the triangle into two halves right

[merry python]

area = (1/s)*b*h as well has use heron's formula

[urban obsidian]

sometimes it doesn't? So how do we know when and when it doesn't

[merry python]

only in case of isosceles and equilateral

[urban obsidian]

oh

wow i didn't know that

thanks

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[unique pebble]

Find all primes a, b, c that satisfy: (a+1)(b+2)(c+3)=4abc

[unique pebble]

thanks for helping

[paper depot]

any progress?

[hasty pulsar]

yw

[paper depot]

ok fucktalogist YOU go ahead and take over.

[unique pebble]

no 😢

[paper depot]

i'd start by considering some cases based on which of a, b and c are 2 and which are not

[vocal sleet]

@unique pebble Has your question been resolved?

[unique pebble]

anyone here..?

[merry python]

Yes

I am on it

solved one case

2nd case is remaining

Okay, so first of all expand it

expand the LHS

[unique pebble]

sure

(a+1)(b+2)(c+3)=4abc<=> abc+bc+2ac+2c+3ab+3b+6a+6=4abc

<=> bc+2ac+2c+3ab+3b+6a+6 = 3abc

[merry python]

Okay now send all terms divisible by 3 to the RHS

[unique pebble]

bc + 2ac + 2c = 3abc-3ab-3b-6a-6

[merry python]

So you can conclude that bc+2ac+2c=3k

[unique pebble]

so bc+2ac+2c divisible by 3

[merry python]

now factor out the c

you are left with two cases

[unique pebble]

c * (b+2a+2) divisible by 3

[merry python]

Yes

[unique pebble]

case 1: c =3 then a=5, b = 3

[merry python]

Yes

case 2 is where I'm currently working on

[vocal sleet]

@unique pebble Has your question been resolved?

[unique pebble]

<@&286206848099549185>

[vast shale]

how can I help you ?

[cobalt shard]

Question?

[boreal condor]

Hello

[unique pebble]

here

[cobalt shard]

(ab+2a+b+2(c+3)=abc+2ac+bc+2c+3ab+6a+3b+6=4abc

[boreal condor]

Where did you find this monstrosity

[cobalt shard]

Therefore 2ac+bc+2c+3ab+6a+3b+6=3abc

[boreal condor]

Homework?

[unique pebble]

yes

[boreal condor]

What level of maths is this

Just to prepare myself

[unique pebble]

9 grade math

=))

[cobalt shard]

Huh

[boreal condor]

Ah as I expected

[cobalt shard]

No

You’re joking lmao

What is it actually?

[unique pebble]

that's true

I'm an Asian so every Math problems are Satan

[cobalt shard]

Ohh

Lmao

[boreal condor]

Tell me if you finish this question (part of your phd)

[cobalt shard]

Lemme see if my knowledge allows me to do this

[boreal condor]

Maybe there's a trick to it

ab + 2a + 2b + 2c + 2 = abc?

[unique pebble]

what

[boreal condor]

Let me verify one sec

I may have rushed it

[unique pebble]

I expanded everything here

[vast shale]

This was the right way right

c = 3

Then check for others by trial and error?

Now , (a+1)(b+2)(3+3) = 4(ab)(3)

(a+1)(b+2) = 2ab

Yeah 5 and 3

[vocal sleet]

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[vocal sleet]

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[kind shore]

how do i do this?

[kind shore]

i did the rational root theorem

and wrote out al possible roots

but im confused what to do from here

[formal pond]

am i trippin?

could u not just pick the ones with i's?

[crimson jetty]

Factor theorem

P(r) = 0 iff r is a root

[kind shore]

nah i already tried that

[kind shore]

ohh i see

[crimson jetty]

idk what the rational root theorem is

presumably the same result

[kind shore]

but i thought complex roots

had to have i in it

[crimson jetty]

0 is a complex number.

as is 1/5

[kind shore]

ah

so

do i just have to start plugging in numbers

and

[crimson jetty]

well yes if u are to use this result

actually that isn't even the factor theorem

[formal pond]

that works too

[crimson jetty]

that's just the definition of a root

[spiral inlet]

all the rational root theorem would do here is rule out some of these possibilities, you'd still have to plug in the rest to check

[kind shore]

is there an easier way?

[crimson jetty]

plugging in should be easy enough until you get to the non-real ones 🙄

actually no, there are no mixed real-imaginary complex numbers

all the things they want you to try are real or purely imaginary

shouldnt be much pain to plug in

[vast shale]

Are you a complex number

xD

[kind shore]

oh man this is gonna take a while

auhgauhgduh

[vast shale]

this is v easy to do acc dw

First see that the y intercept is 16

So x=0 can be crossed out obv

[kind shore]

mhm

[vast shale]

3* 32-6 *16+11 *8-22 *4+ 8 *2-16 notice sth about this?

3* 32 is the same as 6* 16

11* 8 is the same as 22* 4

So x=2 will give 0

That gives u one root easily

[kind shore]

wait where did we get these numbers

[vast shale]

So select x=2

[kind shore]

oh

plugged in 2

[vast shale]

Yeah

Now you just need one more then you've got a quadratic which you can solve easily

If it's complex you can cut down 2

So try the easiest one first

+i

3i - 6 - 11i + 22 + 8i - 16

[formal pond]

WAITT

[kind shore]

huh

[vast shale]

Substitute x=i

[kind shore]

i see

[vast shale]

That's not zero

[formal pond]

Yea sub would be the easier

[vast shale]

So x=-i is gone too

You know this is gonna have at least 2 complex roots so just without thinking select x= ± 2i sqrt(6)/3

[kind shore]

i see

[vast shale]

You know bcz in the options you removed 2 options already

Now you can write p(x) as this

p(x) = (x-2)(x^2+8/3)q(x)

q(x) will be a quadratic

[kind shore]

hm

[vast shale]

You can now guess x=1/5 as another root since there's no other option left

Or you could try to multiply (x-2)(x^2+8/3) and then quotient it w the main polynomial to figure out q(x) then find it's roots

But that's too long n it's a mcqs for a reason

So just select x=1/5 and get done w it

Done

[kind shore]

ooh i see

thank you!

.close

[vocal sleet]

Channel closed

Closed by @kind shore

Use .reopen if this was a mistake.

[vocal sleet]

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Remember:
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[viral copper]

what do you mean by order

[hard atlas]

order in which sense?

[viral copper]

Do you mean degree of a polynomial function

[hard atlas]

then yes

[viral copper]

yes the degree of the derivative is always n - 1

[hard atlas]

(except constant polynomials)

well and the derivative doesnt have degree -1

[vocal sleet]

Channel closed

Closed by @unreal pier

Use .reopen if this was a mistake.

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Send your question here to claim the channel.

Remember:
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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.

[acoustic flower]

hui

[acoustic flower]

hi

i need help

i dont understand ratio and proportions

i did in 7th grade but i have to know it again

i dont understand T-T

[obtuse sierra]

https://www.khanacademy.org/math/in-class-7-math-foundation/xe6a68b2010f94f8c:ratio-and-proportion

[small quarry]

ratio, is like fractions

let me use the example of Khan Academy basic ratios Q.1

apple is 3

total of fruit is 8+5+3 = 16

apples to total pieces of fruit

Therefore: 3 to 16

[acoustic flower]

ok ok that i understand but

i dont understand the proportions in Problems