help-49

[lyric charm]

the woman's plan of action is to try keys one by one, discarding the wrong ones, until she finds one that works

[twilit field]

yes

[lyric charm]

i would argue it makes tons more sense to consider the sample space here to consist of all permutations of all n keys

corresponding to the order in which they're drawn from her keychain

[viscid jolt]

hmm

[lyric charm]

succeeding on the k'th try now means that the good key occupies the k'th spot in the permutation

while the others can be arranged in whatever way

[twilit field]

so I have to permute (n-1) keys

[lyric charm]

if you wanna phrase it that way, sure.

[twilit field]

Which can be done in (n-1)! ways

and the size of the sample space would be n!?

[lyric charm]

yes

[twilit field]

and there are k ways to place the right key

[lyric charm]

no

there is only one

the good key has to go in exactly the k'th spot

it has nowhere else to go, bc it has to be drawn on exactly the k'th try and at no other point

[twilit field]

so the answer would simply be (n-1)!/n!=1/n

[lyric charm]

yes

[twilit field]

Same question, but she doesn't discard the previously tried keys this time

[fallen sparrow]

interesting question

[fallen sparrow]

ye

[lyric charm]

are you at the stage where the words "independent events" have appeared

(yes or no, no wrong answer except if you decided to lie to me)

[twilit field]

That's the next chapter

[lyric charm]

mkay

[fallen sparrow]

youre studying probability?

[lyric charm]

thing is though now defining the sample space and the probability measure is going to be tricky if we're doing it in full detail

(bc the girl might take arbitrarily long to find the damn key)

[twilit field]

yes, revising basic probability before we start random variables

[twilit field]

I'll check how much detail is required with my prof tomorrow

[lyric charm]

but the idea is still that the first k-1 tries have to fail and the k'th has to succeed

[twilit field]

yes

[lyric charm]

i guess maybe we could now cap ourselves to k attempts and consider that either the girl gives up if she doesn't get it on the k'th try or that we don't care what happens beyond the k'th try anyway

[twilit field]

so the probability would be (n-1)^k/n^k

[lyric charm]

(n-1)^(k-1) / n^k

[fallen sparrow]

what ann said

[twilit field]

right

okay

thanks

[molten bay]

Can i join the party?

[hard shard]

heres a simple alternative: shuffle the keys like a deck of cards, what is the probability key k is the right k?

[lyric charm]

what party?

[hard shard]

right key*

[lyric charm]

he's now doing it with a reshuffle after each draw

[hard shard]

i mean, i guess

but that doesnt affect the probability at all

if you imagine she tries each key anyway even after getting the right key, you get a sequence

(at the end)

[twilit field]

hmm okay

[hard shard]

the probability of key k being right is independent of the other keys

[twilit field]

Another question

uh

, wait a bit

thanks everyone

.close

[midnight plank]

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[midnight plank]

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[willow spindle]

if $\alpha$ and $\beta$ are the roots of $x^2+x+1$ and $\gamma$, $\delta$ are the roots of $x^2+3x+1=0$ then value of $\alpha^{\delta}+\beta^{\gamma}+\alpha^{\gamma}+\beta^{\delta}$ is

[grand pond]

Madhup

[willow spindle]

Pls help guys

[grim vector]

!msgdel

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[barren wharf]

Find integer solutions to:

x^(2y) + xy^15 - x! + y!/(5!) + 2^x + x = 7952735876771577277868825991

[jaunty ivy]

what have u tried

[barren wharf]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[barren wharf]

,w factor 7952735876771577277868825991

[grand pond]

Results provided by WolframAlpha

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

Requested by normie12345678910111213

[barren wharf]

Interesting.

[thin relic]

$$ $x^{2y}$ + $xy^{15}$ - x! + $\frac{y!}{5!}$ +$ 2^x$ + x = 7952735876771577277868822995 $$

[jaunty ivy]

where did u get this question

[grand pond]

ALPHA NEEL

$$ $x^{2y}$  + $xy^{15}$ - x! + $\frac{y!}{5!}$ +$ 2^x$ + x = 7952735876771577277868822995 $$
```Compilation error:```! Display math should end with $$.
<to be read again> 
                   x
l.49 $$ $x
          ^{2y}$  + $xy^{15}$ - x! + $\frac{y!}{5!}$ +$ 2^x$ + x = 795273587...
The `$' that I just saw supposedly matches a previous `$$'.
So I shall assume that you typed `$$' both times.```

[barren wharf]

My friend.

,w x^(2y) + xy^15 - x! + y!/(5!) + 2^x + x = 7952735876771577277868825991

[grand pond]

Results provided by WolframAlpha

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

Requested by normie12345678910111213

[austere blade]

this question looks awfully hard though

judging from the massive number

[jaunty ivy]

its a troll

[barren wharf]

?

Just because you can't do it does not mean I am a troll.

[thin relic]

$x^{2y}$ + $xy^{15}$ - x! + $\frac{y!}{5!}$ +$ 2^x$ + x = 7952735876771577277868822995

[grand pond]

ALPHA NEEL

[thin relic]

This is it

[tawdry laurel]

Where is this question from

[barren wharf]

...

[tawdry laurel]

Have you tried plotting on a 3D graph?

[barren wharf]

Nope.

[austere blade]

did your friend conjure it up out of nowhere 😭

[tawdry laurel]

Try it

[barren wharf]

how

[tawdry laurel]

You can get triples of solutions

[tawdry laurel]

Desmos

[barren wharf]

What is the third variable

[tawdry laurel]

anything

[jaunty ivy]

even if u could bound this, u'd have far too many testcases for it to be a reasonable question.

[tawdry laurel]

just graph it lmao

[barren wharf]

okay

I get sheets of metal.

<@&286206848099549185>

[snow dawn]

Ok

Time to help my name sake

[snow dawn]

Looks like my namesake is also preserving my reputation as a troll

Neat.

Yeet

5 factorial is 120

Multiply that mass of number at rhs by 120 and solve for x and y the exponential way using exponential laws if needed

Also idk why you would ever need or get to solve this kind of question @barren wharf

Just skip it

It adds no value to your life even if it comes to exam

Solving it would most likely make you fail the exam

Also do I know you?I have so many stalkers irl who make up these ids and come to my servers with intentions I don't want to talk about 👀

[midnight plank]

@barren wharf Has your question been resolved?

[midnight plank]

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[primal owl]

this is a question I've been sort of avoiding for a while and i wanna get to it today :p
the first part (finding the x coordinate of the stationary point) was simple enough (x=pi/12). the derivation function turned out to be dy/dx = (3e^3x)(sin3x-cos3x)/sin²3x

idk how to go about finding the second derivative tbh it just looks so tedious 😭 is there any other way to determine the nature of the stationary point?

[primal owl]

i tried to analyze the original function and found that e^3x is greater than zero in the interval (and beyond that obviously) and the sin(3x) is negative between [pi/3, pi/2) but idk if I can use that info :(

in my mind since pi/12 is less than pi/3 then it should be a maximum point but idk

[austere blade]

I don't think there's a way to dertimine the nature of the stationary point without doing the second derivative test

[primal owl]

aw :(

idk if my exam board would give a question THAT long though so i assumed there might be a clever way to do it :p

[austere blade]

there might be but I can't really see if there's a trick to it

[austere blade]

what exam board is it

[primal owl]

cambridge international examinations (basically js a levels)

[austere blade]

Yeah they can be mean with it when it comes to questions like these

[primal owl]

i am NOT differentiating that thing tho if they expect me to do that 😰

[austere blade]

What's the marks out of, (if it displays it)

[primal owl]

it doesnt, its a question from my book and they dont show that there (which is recommended by my exam board so its supposed to be a primary resource apart from past papers)

BASICALLY IF YOU CAN GET THE TEXTBOOK QUESTIONS u can move onto past papers

and i cant get this one so im nervous abt it

i left it 2 days ago but its been lingering in my head :D

[austere blade]

The textbook questions usually have questions that require quite long working out

[primal owl]

yeah im aware some can acc be more difficult than actual exam question

i guess this is one of them

[austere blade]

It is quotient rule again but a bit more tedious 😭

[primal owl]

lemme js try finding the second derivative again atp

[primal owl]

DO THEY EVEN GIVE ENOUGH SPACE FOR ALL THAT 🥀

ok im procrastinating lemme js do it

[austere blade]

alright, good luck with it

[midnight plank]

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[snow dawn]

3^x = x⁹. Solve for x

[midnight plank]

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

[snow dawn]

I think we need to make 3 and x equal in some way

Which is not possible

I think this question more suitable somewhere else

.close

[midnight plank]

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[small jasper]

lambert function time

[snow dawn]

.reopen

[midnight plank]

✅

[snow dawn]

Nvm

How do we use Lambert W

<@&286206848099549185>

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[snow dawn]

Quelle?

This is my first ping

[hushed mauve]

it's only been 5 mins

[lusty furnace]

W(xe^x) = x

[snow dawn]

Oh I see sorry

What is e? Euler?

[summer geode]

yes

[lusty furnace]

Euler’s number yes

[snow dawn]

What does W do then

[lusty furnace]

Reminder that $a^b = e^{b \ln(a)}$

[grand pond]

Médicis

[summer geode]

W is the inverse of $f(x) = xe^x$

[grand pond]

Yeatte

[summer geode]

so $W(f(x)) = x = W(xe^x) = x$

[grand pond]

Yeatte

[summer geode]

might be some extra conditions

[snow dawn]

Let's see

Is W = a^i?

[summer geode]

it's a multivalued function so inverse might not be the best word

[visual tiger]

W(xe^x) = x is not always true, depending on which branch of lambert function you take

[summer geode]

^

[snow dawn]

Actually its

we^w = x

[summer geode]

w isn

w isn't a number

https://en.wikipedia.org/wiki/Lambert_W_function

[snow dawn]

It's a complex number?

[summer geode]

no, its a multivalued function

[visual tiger]

it's true though that W(x)e^W(x) = x

[snow dawn]

So tell me the steps to solve the question please

[summer geode]

you didn't post a question to solve?

[snow dawn]

.

[visual tiger]

this question

[summer geode]

oh the previous thing before this?

mm

[visual tiger]

this one actually you can possibly solve without lambert

suppose x = 3^t

(x has to be positive because x^9 = 3^x is)

and solve for t

[snow dawn]

3^(t)9 = 3^3^t

[visual tiger]

mmmh well even after simplifying this is a bit sad

you gotta "spot" the solution basically

[snow dawn]

I.e. 3^9t = 3^3^t

So 9t = 3^t

I guess we raise both sides to the power of 1/(t)

[visual tiger]

ehhhh

not really

this is all about guesswork now, or lambert

[mighty flicker]

maybe consider log to base 3?

[snow dawn]

What are we guessing

[visual tiger]

doesn't change the problem

[visual tiger]

guess for a t that works

9t = 3^t

you can divide by 9 if you want and that changes to t = 3^(t-2)

[snow dawn]

Trial and error method 🙆

Why I chose to raise both sides to 1/(t)

[mighty flicker]

wait i think im cooking something

[snow dawn]

Was to get rid of the exponent t on 3

We get a more simpler equation

[visual tiger]

what's that

[snow dawn]

(9t)^1/(t) = 3

Now we have to try and eliminate one t

[visual tiger]

put in spoilers at least please

[snow dawn]

Please don't give answers 🙂

[mighty flicker]

oopsies

welp dm me if you want the method

or just ping

[snow dawn]

Just write it here for Pete's sake

Actually my question now is

[mighty flicker]

alr then

[snow dawn]

How do we eliminate 1 t from the LHS

[visual tiger]

you can't, not really

[snow dawn]

Then we just have to find a value of t that satisfied that equation

[visual tiger]

also, while the first solution is obvious, there is another... and can't be expressed without lambert W

[snow dawn]

Which is not a method in its own right

[visual tiger]

but you're tasked to solve for all solutions in the equation you posted

one of them literally can't be given in exact expression without W lambert function

[visual tiger]

anyways if you want to guess the first solution I suggest you look at this

[snow dawn]

You said to find a value of t

[visual tiger]

or this

[snow dawn]

If it is to find all solutions

That method can't work

[visual tiger]

the first solution we can write without using lambert W function at the very least

[snow dawn]

But it's a guessing thing with no real way of actually getting there. It will take eternity since I see numbers up to 5 don't satisfy it

[visual tiger]

I'm just you'd rather write your solution as "x = 12" rather than "x = 2^W(1/2)" or something

[visual tiger]

nah you definitely missed it

[snow dawn]

?

[visual tiger]

if you went up to 5

then you miscomputed something

let's do the computations together then

t = 1, what is 3^t, and what is 9t

[mighty flicker]

maybe the way to solve it is to convert it to the simplest form possible and start guessing from there

[visual tiger]

how is 3^t = 9t not the simplest form possible

[snow dawn]

For 1

We get 9 = 3

For 2 we get 4.24 = 3

[visual tiger]

so that's not correct

[mighty flicker]

simplest form would be ||a+2 = 3^a where t = 3^a||

[visual tiger]

4.24?

[snow dawn]

Yes

(9 . 2)^1/(2) =3

Sqrt. 18 = 3

4.24 = 3

[visual tiger]

ok... do that form if you prefer

but computing n-th roots is gonna be bad for you in the long run

what about t = 3

[snow dawn]

Plus

We can't use calculators, we can simply tell that there are no integral roots to the expression simply by looking at it

[visual tiger]

if you can't use calculators

[snow dawn]

Sqroot. Of 18 has not integral roots

[visual tiger]

then WHY are you using the (9t)^1/t = 3 form

instead of 9t = 3^t

[snow dawn]

Because 9t^1/(t)is easier for me

[visual tiger]

ok... if you say so

what about t = 3

[snow dawn]

I am well equipped with the laws of exponentials

Now we can see that

[visual tiger]

it's (9t)^1/t

not 9t^1/t

jic

[snow dawn]

True that

(9 . 3)^ 1/3 = 3 is actually a true equation

[visual tiger]

that you missed the first time

[snow dawn]

The cube root of 27 is 3

[visual tiger]

ok

then you can notice that the bigger t is from there on, the smaller (9t)^1/t is gonna be

[snow dawn]

So that means that t is 3

[visual tiger]

t = 3 is the only "nice" solution yes

[snow dawn]

That means x =3³ = 27

Then how do we find the other sol.

It will be a long journey with no fruitful results

[visual tiger]

either expressing through lambert W function, or we can just prove it exists and is the only other solution

[snow dawn]

But thanks 🙂

[visual tiger]

which one you prefer

[snow dawn]

Lambert W affcorse

[visual tiger]

ok

well

9t = 3^t

is the same as $9t = e^{t\ln 3}$

[grand pond]

Raphaelisius Maximus MMIII

[digital juniper]

i am suffring to a be an average in math
so help me to solve this

[visual tiger]

I don't know if you use ln or log for natural log

[visual tiger]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[visual tiger]

so

let's just write everything together

[snow dawn]

Natural log is definitely ln

[visual tiger]

$9te^{-t\ln 3} = 1$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

now we try to make some "Xe^X" appear

the exponent can't really be changed

[snow dawn]

Wait did you just transpose the exponential by just making it negative?

[visual tiger]

so we'll make the multiplicative coeff match

[visual tiger]

dividing by e^(...) is the same as multiplying by e^(-...)

exponent rules

[snow dawn]

Oh yes the e got multiplied to 9t as well didn't see

[visual tiger]

now, we want the left hand side

to be Xe^X

[snow dawn]

I don't see any W here

What's the issue

[visual tiger]

you'll see when we start needing it

[snow dawn]

Ok

[visual tiger]

remember that Xe^X = a means that X = W(a)

and that's when it'll inevitably appear

make the multiplicative coefficient match the exponent:

$-t\ln 3 e^{-t\ln 3} = -\frac{\ln 3}{9}$

[grand pond]

Raphaelisius Maximus MMIII

[snow dawn]

So W is the name of the function of a. When a is x, then?

[visual tiger]

?

[visual tiger]

we have Xe^X = ... with X = -tln(3)

[snow dawn]

I am trying to understand this

[lusty furnace]

W is the reciprocal function of xe^x

[snow dawn]

Does this mean

That we replace a with x

Or something else it means

Which implies, Xe^X = X

[visual tiger]

no?

[lusty furnace]

a reciprocal function is named $f^{-1}$ and is defined as $f^{-1}(f(x)) = x$

[grand pond]

Médicis

[visual tiger]

a is some expression

if you apply "xe^x" to X and you get a

that means that if you "undo xe^x" to a, you get back X

that's what an inverse function is

[snow dawn]

Forgive me I have never been taught Lambert function 🙏

[visual tiger]

think of it just as an inverse function

take some positive number a

and now I have a mystery number x, positive

such that x^2 = a

how do I find x

[snow dawn]

I think I can use my first method

Directly in this question

So the question is 3^x = x⁹

Can we raise both sides to the power of 1/(9) to isolate x?

[visual tiger]

if you want to?

[snow dawn]

We get, (3)^x/(9) = x

Now we have to separate variables

Such that x and x both come to the left hand

How do we do that though

[visual tiger]

you can have both x's on the same side if you want

but you won't be able to "separate" them

[snow dawn]

Texit this pls

[visual tiger]

$3^{\frac x9} = x$

$1 = x3^{-\frac x9}$

[grand pond]

Raphaelisius Maximus MMIII

[snow dawn]

Now we have to make x single

Not possible I guess

[visual tiger]

no

[snow dawn]

Your method serves it better

[visual tiger]

lambert is almost always the only way in those problems

if you have some number x, in terms of its exponential

it's almost always lambert

no way around it but guesswork (which doesn't give you all the solutions)

[snow dawn]

Wait

The next step should be

This

Hmm

Something is in my mind

[tropic ice]

I haven't read the whole conversation but if you want on YouTube there is a channel called Prime Newtons that explains the W function in a very simple way, at least I learned about it for the first time through that channel so in case I recommend it to you

[snow dawn]

Sure

Maybe

[visual tiger]

I can let you think about it if you want, but I don't think it will be that promising

[snow dawn]

I can further simplify this to

By raising both sides to the power of 1/x

We get that (3)^1/(9) = x^1/(x)

We got both x on the rhs

We can rewrite them to the LHS for convenience

Now it looks pretty straightforward doesn't it?

[visual tiger]

how do you propose we move forward then

[snow dawn]

We think

Let us see

3 and x are not the same bases

So we can't assert that the exponents are equivalent to each toher

So we have to make 3 and x equal to each other as I stated earlier

Which can't work because x and 3 were never meant to be equal and x = 3 doesn't satisfy the equation

I am cooking sumthing

There is a method I normally use in integration

I call it the blindfold method

We simply make two powers that when they multiply each other they give 1. So raising the equation to those powers will still give us the same equation, but it gives room for manipulation

Technically this is allowed since the both equations are equivalent

We use this whenever we want to "manipulate" something like you get what I am saying

So let's take any power a

Let's use 3 for convenience

3 × 1/(3) = 1

These powers can be used for manipulation

[mighty flicker]

but you did that manipulation with the answer in your mind tho

[snow dawn]

Normally what I do in integration is that this gives me a way to you know, get it in a way that it actually allows me to use one of the laws

And I arrive at the solution

This is called manipulation

[visual tiger]

I doubt your manipulation will get you anywhere but you're free to try. I do feel like you would be wasting your time tho

[snow dawn]

Here the only applicable law is (a^m)^n = a^m.n

[snow dawn]

No bro

Our problem is that we can't find all solutions

So I am deriving a technique that solves it as an algebraic equation

Which will obviously give us all real solutions

Here

If we try to simplify it then we arrive back at where we started because 3× 1/(3) = 1

So here is an actual smart method I can use by mere looking at this

How about swap positions of powers 1/9 with 3? This will mean that 1/9 × 3 will no longer give us 1. Hence, manipulated

😈

How can I do this?

Commutative property of multiplication

A×B = B×A

Upon raising the LHS to (((3)^1/(9))^3)^1/(3) = x^1/(x)

We get through Commutative property that

(((3)^3)^1/(9)^1/(3) = x^1/(x)

[snow dawn]

Texit this if you want to

This can be simplified to: 27^1/(27) = x^1/(x)

The answer is crystal clear

For the RHS to be equal equal the LHS, x= 27. And this is one answer to the question

Maybe using different powers will give us other answers Idk I haven't checked

But I am happy this worked 🙂

Or maybe this is the only real solution to the problem

Because any other power will still be equals to 1

Not changing the skeleton of the equation arriving us at the same answer

So x = 27 is the only real solution to this equation, algebraicly

Anyways...

I have a more intense question

In question 7 I struggle to draw the correct graph needed to solve for the equation

I was marked incorrect for some reason even upon shading the region that had values in common

I don't understand why

[shadow schooner]

,rccw

[snow dawn]

Here is my working

Here is the graph I drew

Why is the region shaded the incorrect requirement for the solution?

[fallow scarab]

This channel is too long to catch up. Should open a new channel with a summary of your question and progress

[snow dawn]

It starts from here

[shadow schooner]

yeah new question new channel could be a good idea nonetheless

[snow dawn]

Not too long

K

.close

[midnight plank]

Channel closed

Closed by @snow dawn

Use .reopen if this was a mistake.

[visual tiger]

no it isn't :(

it's just the only nice one as I stated

[snow dawn]

You once stated that W is not a number @visual tiger

[visual tiger]

W is a function

it takes numbers as inputs

and outputs other numbers

[visual tiger]

just like $\sqrt$ is not a number, it's a function

[grand pond]

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

[visual tiger]

it takes inputs such as 4 and outputs 2 in that instance

well

W takes inputs

and outputs stuff too

[snow dawn]

Oh I see

So W is the inverse function of xe^x?

[visual tiger]

so... there is another solution, a number, that can be expressed using the lambert W function

[visual tiger]

yes

in a way

just like sqrt(a) chooses to output the positive number such that x^2 = a

there are two branches to W lambert function

$W_0$ and $W_{-1}$

[grand pond]

Raphaelisius Maximus MMIII

[snow dawn]

Why specifically 0 and -1?

.reopen

[midnight plank]

✅

[visual tiger]

it's notations, I don't remember why exactly

I think it's because the lambert function breaks at -1

as -1e^(-1) = -1/e, and it's the turning point of the xe^x function

$W_0(a)$ is the only $x\geq -1$ such that $xe^x = a$

$W_{-1}(a)$ is the only $x\leq -1$ such that $xe^x = a$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

and just like we need $a\geq 0$ for $\sqrt a$ to even exist

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

we need $a\geq -\frac 1e$ for $W_0(a)$ to exist

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

and we need $-\frac 1e\leq a < 0$ for $W_{-1}(a)$ to exist

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

W lambert function is much nastier as an inverse function than the square root in that sense

but it still fulfills its role as an inverse function

so, just to quickly come back to your problem

[snow dawn]

I see that 0 is the minimum value and -1 is the maximum value. They are just connotations

[visual tiger]

so we can finish this actually knowing where the second solution comes from

[visual tiger]

probably this is a good explanation

so

$3^x = x^9$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

we can rewrite as $x3^{-\frac x9} = 1$ as we did previously

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

$xe^{-\frac x9 \ln 3} = 1$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

we want (something)e^(same thing) = ...

so we can apply lambert W

so

$-\frac x9 \ln 3e^{-\frac x9 \ln 3} = -\frac{\ln 3}{9}$

[grand pond]

Raphaelisius Maximus MMIII

[snow dawn]

Where is the Lambert W .-.

[visual tiger]

$-\frac 1e \leq -\frac{\ln 3}{9} < 0$, so there will be two solutions

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

name $X = -\frac x9 \ln 3$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

then $Xe^X = -\frac{\ln 3}{9}$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

so $X = W_0\left(-\frac{\ln 3}{9}\right)$ or $X = W_{-1}\left(-\frac{\ln 3}{9}\right)$

[grand pond]

Raphaelisius Maximus MMIII

[snow dawn]

The question arises

How did you come up with -ln 3/(9)?

[visual tiger]

we start from this

we remember we want Xe^X = something

to apply W lambert

[visual tiger]

so let X = the exponent

and now rearrange by multiplying or dividing

to get Xe^X = ...

[visual tiger]

that's what it looks like afterwards

[visual tiger]

now the solutions that we get from both branches of W

and now solve back to x

$-\frac x9 \ln 3 = W_0\left(-\frac{\ln 3}{9}\right)$ or $-\frac x9 \ln 3 = W_{-1}\left(-\frac{\ln 3}{9}\right)$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

$x = -\frac{9}{\ln 3}W_0\left(-\frac{\ln 3}{9}\right)$ or $x = -\frac{9}{\ln 3}W_{-1}\left(-\frac{\ln 3}{9}\right)$

[grand pond]

Raphaelisius Maximus MMIII

[snow dawn]

Sorry I have gotten a bit confused

How did you change 3^x = x⁹ into this?

[visual tiger]

^1/9

then divide by 3^(x/9)

which is the same as multiplying by 3^(-x/9)

[snow dawn]

Oh

Does this always give you an expression like xe^(something), like almost always?

[visual tiger]

that's what we're looking for by design

if you have e^(....) = x or something

you can simply put the exponential to the other side

knowing that it will just carry a minus in the exponent

and so you will have (something1)e^(something2) = ...

[snow dawn]

I am guessing this won't be the same for 3^x = 2^2x

For this we still have to make it shape to that design

[visual tiger]

this is a different type of equation

and actually is easier

lambert applies mostly to "x^... = ...^x" types of equations

so x acts as both the base of one term and the exponent of another

[visual tiger]

here, x only acts within the exponents

and so this is a matter to be left to logs

[snow dawn]

Oh true

[visual tiger]

$x\ln 3 = 2x\ln 2$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

you'll also find that this is only true for x = 0

[snow dawn]

Ok thank you, I appreciate this lecture 🙂

!done

[midnight plank]

If you are done with this channel, please mark your problem as solved by typing .close

[snow dawn]

.solved

[midnight plank]

Channel closed

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[midnight plank]

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[stark verge]

The quadrilateral ABCD is a parallelogram. The segments AE, BE, CG, and DG intersect the interior angles of the parallelogram.
Prove: Angle BFC=90. I didnt manage to solve this after hours of work.

[shell wigeon]

Do you mean "bisect the interior angles"?

[stark verge]

like ADH HDC are equals FCD and FCB equals etc

[shell wigeon]

Yeah, bisect

As in "split in two equal parts"

Anyway, AB and CD are parallel, so what's the sum of angles ABF, FBC, BCF, FCD?

[stark verge]

I'm assuming 180, right? Because in a parallelogram, adjacent angles are complementary to 180.

[zenith snow]

And in triangle BFC,the sum of all three angles
<BFC+<FBC+<BCF=?

[shell wigeon]

Yes

ABF = FBC and BCF = FCD, as given by the bisectors, so you should be able to find FBC + BCF, and then BFC

[midnight plank]

@stark verge Has your question been resolved?

[stark verge]

thanks

[mortal mirage]

@stark verge you still need help?

[midnight plank]

@stark verge Has your question been resolved?

[midnight plank]

Channel closed

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[midnight plank]

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

statistics

can someone pls tell me why #24 is wrong

26/90 is 0.28888888889

rounded to 2 decimal places should be 0.3

why does this website say its wrong

RF relative frequency is frequency over total so i put 26 over 90

[past jay]

Hmm

I'm sure the total amount of men is 99

This might have been where you made your mistake

[calm shuttle]

ohh wait i just calculated it again

And ur right thank u

😭

[past jay]

No worries

[calm shuttle]

but wouldnt the answer still be 0.3?

because if i divide 26 by 99 now i get 0.26

[past jay]

2 decimal places

So you don't write 0.3

You write 0.26

[calm shuttle]

Wait im so confused

Why did 0.4 work

For the last one

Even if it was 1 decimal place

And do i not have to round it?

[past jay]

Because it coincidentally is 40/99 = 0.404...

So 2 decimal places is 0.40

[calm shuttle]

Ohhh

Bye im like slow

My bad

Thank u tho

[past jay]

Nw

[nimble leaf]

technically you should write 0.40

[midnight plank]

@calm shuttle Has your question been resolved?

[midnight plank]

Channel closed

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

a person wrote out 10.5^47 as an answer in full as a decimal number, how many digits did she write

[mortal falcon]

hint: 10.5 = (10 + 1/2)

hint 2: ||binomial theorem||

[fierce canyon]

we havent learnt about that yet

[mortal falcon]

okay you can also do this with logs

if you have seen those?

[fierce canyon]

havent done either 😭

[mortal falcon]

im stumped i think someone else is able to help though

[fierce canyon]

hopefully

[mortal falcon]

what unit is this for?

[fierce canyon]

its like a test for all maths

from a past paper

[modern sapphire]

the number of digits beyond the decimal points is gonna be the same as the those in 0.5^47, so deffo 47 beyond the decimal point, and 48 before

[feral sedge]

it's not exactly obvious that 1.05^47 is less than 10 (indeed it is 9.90...)

[fierce canyon]

can you explain a bit

[modern sapphire]

0.5^47 is same as (5/10)^47

and 5^47 doesnt have any decimal places

so, the whole number gets shifted by 47 places once divided by 10^47

[modern sapphire]

the number is 10.5, not 1.05 in op's question

[feral sedge]

?

[fierce canyon]

yeah but how do we know how many digits 5^47 has

[feral sedge]

10.5^47 = 10^47 * 1.05^47

so obviously the integer part has either 48 or 49 digits

but how do you know it is 48 for sure

[modern sapphire]

oh ic

[feral sedge]

it is extremely unclear

at least not without some more specific analysis

[fierce canyon]

how do we find the digits in 1.05^47?

[modern sapphire]

I may have used something I know about 1.1^48, tho you are right, I would need to establish that

[modern sapphire]

we dont. and tbh we dont care, coz no matter what its gonna be less than 47, and the point is to find the number of digits after the decimal point, and not before

we can simply prove that the last digit of 5^47 (which is def not 0, coz there is no factor of 2 there) gets moved 47 places after the decimal point

and thats the only contributor to the smallest digit

[fierce canyon]

how do we find the amount of digits after the decimal point

[modern sapphire]

10^-47 is easy to find no?

[fierce canyon]

isnt it 5^47 srry im kinda slow

[modern sapphire]

I am talking about $\left(\frac{5}{10}\right) ^{47}$

[grand pond]

Bacter14Fr0g

[fierce canyon]

so 5^47/10^47

[modern sapphire]

5^47 has only 0s in decimal places

its an integer

so, upon dividing by 10 once, it has 1 non-zero decimal place digit

[fierce canyon]

wdym 5 power doesnt have any 0s right

[modern sapphire]

once you do it 47 times, that means, it is gonna shift the decimal point 47 places, so that it ends up having 47 digits in its decimal form, beyond the decimal point

[modern sapphire]

it is an integer

no fraction part

nothing to the right of the decimal point

nothing except zeros

it is 710542735760100185871124267578125.000000000000...

[fierce canyon]

oh

yeah

[modern sapphire]

once you divide by 10, it becomes 71054273576010018587112426757812.5

another division by 10, makes it 7105427357601001858711242675781.25

[fierce canyon]

so it has to be 47 digits from the decimal point once its done 47 times

[modern sapphire]

if you do it 47 times, the last digit 5 ends up in a place of value 10^-47 aka at a place 47 digits beyond the decimal point

[fierce canyon]

but even so how do we finish the question?

@modern sapphire

[native prairie]

Can someone help me with this question? It asks for the sum of f(1)+f(2)+f(3)+...+f(n)

[modern sapphire]

yea, if you dont know logarithms or binomial, its gonna be tough

[modern sapphire]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[fierce canyon]

was there something about like 10^47*1.05^47 that might be easier idk tho

@modern sapphire u here rn?

[modern sapphire]

dude, this wasnt even my idea

how am I supposed to help you with that

[midnight plank]

@fierce canyon Has your question been resolved?

[midnight plank]

@fierce canyon Has your question been resolved?

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[last slate]

Can anyone help pls ive been stuck on this got ages

[pearl hull]

Sure

What have you noticed in this question?

@last slate

[last slate]

That its a semi circle and angle bce = 90

[pearl hull]

BEC*

[last slate]

O ye mb😭

But yh its 90

Thats what ive noticed

[pearl hull]

decent ✅

Nice find, next question, what information can you get by knowing EDC = 136°

[last slate]

Would EBC be 180-136? =44

[pearl hull]

YES

[last slate]

O

Ohhhhh

[pearl hull]

wanna try for 3 minutes before the next hint?

[last slate]

So then ECB is 46?

[last slate]

kk

[pearl hull]

Correct ✅

[last slate]

kk ty

So now im guessing its uh

Something to do with the angles around a point theorem?

or two radii?

[pearl hull]

You can actually apply the same technique we did on finding EBC

[last slate]

for which one?

EAB?

[last slate]

Oh wait is it triangle EDC?

[pearl hull]

It’s another quadrilateral

[last slate]

O

So that means

ECB + EAB = 180?

OHHHHHH

SHI

[pearl hull]

Absolutely ✅

[last slate]

SO ITS 134?

AYYY TYSMM

[pearl hull]

wtf, what’s 1B?

[last slate]

a question from hell..😭

[pearl hull]

-# Sorry, I’m an Asian and I’m more curious about the mistakes

[last slate]

Lmfao im indian i get that😭😭

[last slate]

But yeah idek what this one is lmao

[keen saddle]

use a protactor to measure all the angles

[last slate]

😭😭

[hushed mauve]

use protractor
(not drawn accurately)
congratulations, your pentagon now has 1000 degrees total

[keen saddle]

i use instead of \s sometimes

[hushed mauve]

i also use for a reason

anyway

[last slate]

but uh yeah i have no clue where to start i wish i could use a protractor tho lol

[hushed mauve]

i think i remember there's a theorem that relates the exterior angle of a tangent to something

[modern sapphire]

the circle looks pretty round, so tbf you could get away with using a protractor here

[keen saddle]

oh i didnt see that mb
and ill leave, you all continue

[last slate]

Angle formed whena. Tangent touches circumference = 90?

[last slate]

its on a laptop screen…

And plus im on holiday nd didnt bring it😭

[modern sapphire]

it was a joke my guy

[keen saddle]

i think there was a theorem called alternate angle something

[hushed mauve]

nono

[last slate]

lmao ikik

[fleet axle]

i don't get why any others are equal to EGF lol

[hushed mauve]

that only applies if there's a radius

but obviously there's none here

[last slate]

Ohhh alternate segment?

[last slate]

oh

[keen saddle]

maybe that could help

[last slate]

yea

but applying it is the dodgy prt

[modern sapphire]

just look at which arc corresponds to the angle

[last slate]

wym..

[fleet axle]

none?

[modern sapphire]

it does tho?

[fleet axle]

can you explain

angle EGF

[keen saddle]

@fleet axle lets focus on vyaas first

[fleet axle]

vyaas asked the same thing though

[modern sapphire]

the tangent at G, and the chord EG also subtend an angle within the circle. All the angles that subtend the same arc/segment would have the same measure

[last slate]

Oh so its kinda like the two radii theorem? With the same measurements part?

[modern sapphire]

i have no clue about two radii theorem, do you mean inscribed angle theorem?

[last slate]

If you mean the one where a triangle formed with two radii are equal then ye

[modern sapphire]

no, I do not

[fleet axle]

ok i see. by "correspond" you mean "subtend"

[last slate]

oh

[modern sapphire]

however, I do mean this

[last slate]

ahhh okay

So that means both GED and GBE are equal? As bith the chords correspond

[modern sapphire]

GED and GBE dont even subtend the same thing

[fleet axle]

check the order of the letters

[modern sapphire]

maybe you need to check the angles, coz one of the angles is correct, and the other has wrong order of letters

[last slate]

Here i mean

[last slate]

Wym by order?

[last slate]

Yh how does that matter

[modern sapphire]

Angles DEC and CDE are different

[fleet axle]

different orders mean different angles

[last slate]

oh

[last slate]

o

[modern sapphire]

why have you highlighted BGE? it is not part of the angle EGF, we only care about EGF

[last slate]

Because it is connected to angle g?

[thorny solstice]

Hi!

[last slate]

hi

[modern sapphire]

and why does that matter?

[last slate]

uh

[modern sapphire]

why not CGE?

isnt that also connected?

[last slate]

Oh

um

Shit

I acc dont know

[fleet axle]

hello

(this is an active help channel :p)

we agree we're using this theorem right? https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcSvkzpuR17GnWKWjpf35SpKKVx037RycKmtrg&s

so we wanna find two angles that share the same arc (AB in the above pic)

[last slate]

Wait what

Why that one..

[modern sapphire]

no we dont? we need this

[last slate]

yh

[fleet axle]

well that's a consequence of what i posted technically

ig you're thinking of remembering that property separately and applying it, that's fine too

[modern sapphire]

well yeah, but we wanna help OP and not confuse him

[last slate]

Ye

[fleet axle]

well i usually memorize as less as possible

[modern sapphire]

its better to reference to things which make direct sense

[last slate]

Im not u tho

[fleet axle]

i'm not gonna really say one method is better than the other

[fleet axle]

i'm not the only one that does that

[last slate]

wym?

[fleet axle]

and it's really a recommended way to learn math

[last slate]

im getting so confused now im not even kidding-

[modern sapphire]

well, @last slate back to the problem

[last slate]

yh

[fleet axle]

but it doesn't matter which method is better, you can do whatever you prefer

[modern sapphire]

can you use the theorem in this figure?

[last slate]

OH IN BCG

AND GDB

[modern sapphire]

again? why are you looking at BG?

the question asks for EGF

and other angles equal to EGF

[last slate]

Yeah because triangle gcb is equal no?c

Cus of alternate segment

[modern sapphire]

no?

[last slate]

oh wha

[modern sapphire]

GCB becomes equal to BGA

which is once again, not what the problem cares about

[last slate]

oh

uh

[modern sapphire]

you got this angle you need to care about

[last slate]

Would it be ecg?

[fleet axle]

you're correct BCG and GDB are equal

but that's not what we're after

[last slate]

o

[last slate]

yea

[modern sapphire]

yea, this is one correct angle

tho there are more

[last slate]

k

Is it ebg as well?

OHHH I SEE TYSM

[fleet axle]

any more?

[modern sapphire]

is that all the angles?

[last slate]

ye im pretty sure

Cus of the alternate sefment theorem

[modern sapphire]

lol what? how does this confirm that there are no more?

[fleet axle]

we now may wanna consider this theorem

[last slate]

uh

[fleet axle]

or just the same logic you used to find the 2nd angle EBG

there's one more

at least

[last slate]

thought there were 2?

[fleet axle]

well if you can prove there are no more with the alternate segment theorem, maybe

bacter asked how you proved it

if there is no proof, then there could be more

[last slate]

oh

uh

[modern sapphire]

so till now you got

is there something that looks like has been ignored till now?

[last slate]

The angles in a same segment theorem?

[modern sapphire]

yes

[last slate]

ohhh

So that means BDG is also equal

[fleet axle]

B-D-G?

[modern sapphire]

WHY ARE YOU OBSESSED WITH THE SAME WRONG ANGLE?

[fleet axle]

BDG

[last slate]

😭😭 i have no clue

But thats in the same segment tho?

[modern sapphire]

and what segment would that be?

[fleet axle]

you mean the angle that has its "corner" at point D? @last slate

[last slate]

Yhh

[last slate]

Uh

[fleet axle]

and then the two rays go through points B, and G?

[last slate]

Yee

[fleet axle]

so that subtends arc BG, right?

[last slate]

I mean if BCD is equal then since BCD and CGD are in same segment doesnt it make em equal?

[modern sapphire]

I dont think anyone mentioned BCD here at all

[last slate]

..

[modern sapphire]

and also, BCD and CGD are not equal

[fleet axle]

you might be doing this already but make sure you're paying attention to the order of the letters still

[modern sapphire]

they have no reason to be equal given the information in the question

[last slate]

Ye

[last slate]

😭😭 im so confused rn bro

[modern sapphire]

do you sort the letters of the angle in the alphabetical order?

[last slate]

No i put the angle im talkin about in yhe centre

[fleet axle]

(quick summary - the green and brown angles in this pic both "subtend arc GE" and you wanna find another one)

[fleet axle]

✅

[last slate]

ye

[modern sapphire]

so these are angles BCD and CGD you mentioned

[last slate]

Yep

[modern sapphire]

are you sure these are the angles that you meant?

[last slate]

wait lemme check

[modern sapphire]

if so, what segment do they share? pls highlight it in the diagram

[last slate]

Oh

They dont

[modern sapphire]

Well, we earlier confirmed that you got the angles EBG and ECG correctly, can you highlight the segment that they share which makes them equal?

[fleet axle]

(you could've applied the theorem if BC was tangent to the circle for example, you might've been thinking that. it's not tho)

[last slate]

Uh lemme see

[fleet axle]

that looks like a line connecting the "corners" of the two angles but not necessarily a line that they share

[modern sapphire]

sadly, no, that is not the segment

[last slate]

I meant ir shares the segemnt

[last slate]

Oh what how?

[fleet axle]

what segment do the two angles both "subtend"?

[modern sapphire]

in circles, the section of a circle made by a chord is considered a segment. So this is the "segment" that they share

[fleet axle]

^ they both "subtend" GE

[last slate]

Oh

shi so

Does that mean theres only 2 then?

[modern sapphire]

so any angle that is formed with G and E as vertices is gonna be the answer

[fleet axle]

it depends if there are any more angles that subtend GE

[modern sapphire]

still no

[last slate]

I think ive said GCE

[fleet axle]

yes

[modern sapphire]

ECG is same as GCE which you already said

[last slate]

yeah

BED?

[modern sapphire]

did I not just say that the angle needs to have G and E on its arms?

[fleet axle]

spoiler method: ||sometimes math is about finding patterns. the 2 solutions we have are GBE, and GCE. a pattern here is that they both start with G and end with E||

[last slate]

Wait it doesnt have to be a triangle does it?

[modern sapphire]

an angle is not a triangle

[last slate]

O

Shivi forgot that

So tha means

GDE?

[modern sapphire]

yep

[last slate]

Erm

[modern sapphire]

is that all the angles tho?

[last slate]

Either ive pressed the wrong oens

fuck

[modern sapphire]

you have

[last slate]

shit

So minor problem

That i forgot to account for

when you get a question wrong

The system changes a little bit around

😭😭😭😭😭😭

[modern sapphire]

no wonder you were hyperfocused on that angle 😭

you just forgot to tell us that

well, back to the same stuff

[last slate]

😭 fuck mv

mb

[fleet axle]

it's the same problem but the picture is flipped horizontally and letters renamed

[last slate]

😭 yea

[modern sapphire]

are you sure this time it is the same angle? coz its the 3rd time you would be attempting

[last slate]

huh

[modern sapphire]

may be they changed again

[last slate]

Yeah

Im not too sure tbh

[fleet axle]

i mean when they changed the problem. they just flipped the image and renamed letters
(at least that's one way to view it)

[last slate]

Oh

So i can tell that one of em is bcg

[fleet axle]

my point is just that it's the same problem as before

[fleet axle]

yes!

[last slate]

yee

So then the next would be BDG? Because of it subtending b and g

[fleet axle]

yepp

[modern sapphire]

vyaas on fire 🔥

[last slate]

Yeeee

hmmm so if there is another one itll be BEG right?

[fleet axle]

yeah that's it 🎉

[modern sapphire]

indeed

[last slate]

Lets seeeee

pls work lol

EYYYY

[pearl hull]

[fleet axle]

[last slate]

TYSM @pearl hull @fleet axle @modern sapphire YALL THE BEST

ty all again ill close dis now🙏🙏

.close

[midnight plank]

Channel closed

Closed by @echo bronze

Use .reopen if this was a mistake.

[midnight plank]

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[viral dagger]

whats the motivation to this?

[viral dagger]

like they just let stuff be variables and it just so happens that it creates an equation that grts the question?

[lyric charm]

the motivation is that they consider a quantity that can be easily tracked throughout the operations we perform upon these sets

[viral dagger]

the quantity (k-d)-dn?

[lyric charm]

(k-d)-dn is the net change in the set total

ie thats what it works out to be

[viral dagger]

what about the motivation of completing the cycle

is it like just a trial and error thinggy

[lyric charm]

net change = 0

[viral dagger]

is it just making it more favourable since if we didnt complete it we would get like k=d(n+1)+n with k=2^n-1 and thats messier?

ooh wait ok i think i get it now

thank you!

.solved

[midnight plank]

Channel closed

Closed by @viral dagger

Use .reopen if this was a mistake.

[midnight plank]

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.