help-36

[shell condor]

Okay, so that's log((2 + x))/2 - x))

[tranquil pine]

Its not any normal bracket

Its gif

[shell condor]

GIF is the common bracket for Logarithms, it doesn't matter much in calculus (at least integration)

[tranquil pine]

Oh is that so?

[shell condor]

Eh, or I might be wrong, but it won't affect our question

[tranquil pine]

No im asking genuinely cuz i dont know fr

Well if u say so

[shell condor]

I remember seeing some questions with GIF, so it does matter, but not in this particular question

f(a + b - x) can be written as log(2 + x) - log(2 - x)

[tranquil pine]

Then integrate using parts?

[shell condor]

Nope

A faster method

f(x) was log(2 - x) - log(2 + x)

-f(x) would be log(2 + x) - log(2 - x)

[tranquil pine]

So add both

[shell condor]

And that is f(-x)

(Or f(a + b - x))

Now what do we call a function where f(-x) = -f(x)

[tranquil pine]

Odd

[shell condor]

And the integration of an odd function is?

[tranquil pine]

From -a to a is 0

Is that all?

[shell condor]

Yup

[tranquil pine]

Oh

[shell condor]

And here a is 1

[tranquil pine]

Ye

So shud i just ignore the brackets inside log?

[shell condor]

In such cases, where the lower limit is just the negative of the upper limit, we always first check if f(-x) = -f(x)

[lyric obsidian]

[x] DOES make a difference

[shell condor]

I guess we do, because it shouldn't change the fact that it is an odd function, at least in this question, I'm not sure in general

[shell condor]

Yup I realised that, I added a correction later

[tranquil pine]

Ou

Ok thanks a lot

.close

[final saddle]

Channel closed

Closed by @gray mica

Use .reopen if this was a mistake.

[lyric obsidian]

Idt yall understood

.reopen

[final saddle]

✅

[lyric obsidian]

The function isn't defined for x>0, so it isn't odd

e.g. at x=1, log(floor((2-x)/(2+x)))=log(floor(1/3))=log(0) which is undefined

@tranquil pine You have to consider cases based on how the function is defined

[tranquil pine]

Oh

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

.close

[final saddle]

Channel closed

Closed by @gray mica

Use .reopen if this was a mistake.

Available help channel!

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[lapis wedge]

Does this proof look good?

[lapis wedge]

Or is there a more direct way to prove this

[tired walrus]

do you mean disprove

[lapis wedge]

yeah sorry

oh if im disproving

would it be better to prove the negation

not really sure what that would be of this set though

[opal plinth]

Don't you just need a counterexample?

[lapis wedge]

yeah thats true, is the counter example I proposed correct?

[opal plinth]

You didn't though

[lapis wedge]

oh I didnt?

I thought I took an (a,b) in the left hand side

and showed it wasnt in the right hand side

maybe im misunderstanding what counter example means

[rigid scaffold]

A counterexample would be stating explicitly four sets (A_1, A_2, B_1, B_2) for which the equality is false

[lapis wedge]

ohh ok

[opal plinth]

There may be some slack in the definition of a counterexample but to me it's with concrete values

[rigid scaffold]

You would also try to make them as simple as possible, while it's not mandatory

[lapis wedge]

Ok i'll try to constrcut a counter example

[final saddle]

@lapis wedge Has your question been resolved?

Channel closed

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

,rccw

[worldly mesa]

Which one

Also

!status

[final saddle]

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

[vague anchor]

Apologies for absence

3rd one and don't know how to begin

It's multiple correct

[worldly mesa]

I think you are supposed to use the IVT

Or not....

[vague anchor]

IVT?

[worldly mesa]

High school?

[vague anchor]

Btw I'm an 11th grade student

[worldly mesa]

Yeah ok so nvm

[plucky rover]

This just needs Rolles doesn't it

[vague anchor]

⁉️

[plucky rover]

If a continuous function changes sign it is zero at a point in between

[worldly mesa]

High school level

[plucky rover]

I recall having done this during my JEE prep

[solid notch]

yeah you should have been taught rolles

[plucky rover]

Hence me bringing it up

[solid notch]

unless calculus hasnt started at all

[thin cloud]

I'm high schooler and I already know rolles lol

[solid notch]

but then why are you advance

[vague anchor]

I'm just at the quadratic equations

[worldly mesa]

In my country they don't teach calc like uni

[solid notch]

rolles is just general shit formalised in a statement so people doesnt need to think it from scracth

[thin cloud]

yeah it's kinda obvious when you look at it

[vague anchor]

Yeah it's easy to understand i agree

[worldly mesa]

This looks like they wanted to type (x-e)(x-pi) in the denominaror on the right fraction

[plucky rover]

Unlikely

[worldly mesa]

Then when you multiply by (x-e)(x-pi) its a quadratic

[plucky rover]

They have π + e in the options

[solid notch]

yeah there are a few jee question in the chapter quadratic equation which can be done with rolles theorem quickly , so i suggest you can mark this question , do it the conventional method and come back in like 12th 2nd month

[plucky rover]

So they definitely wanted it to be x - π - e

[worldly mesa]

Oh I am stupid

[vague anchor]

But i should develop my problem solving even without it shouldnt I

[worldly mesa]

Just mutiply by (x-e)(x-pi)(x-e-pi)

And you get a quadratic

[solid notch]

its just location of roots simply , but very lenghty otherwise , you also need to understand how you use the tools you have given

[vague anchor]

Hmm
Probably we aren't even given its solution in it's pdf

[solid notch]

quadratic equation can be solved using the completing the square method too , but do we use it generally?
middle term split does the trick most casses and sometimes we have to use quadratic formula

[vague anchor]

Shall I call someone
A good helper
By pinning her🫣

[solid notch]

you know na how you can solve using this?

[vague anchor]

That will truly look as ugly as hell ig

[solid notch]

chapter name

[plucky rover]

Not all problems will have pretty solutions

Especially when you don't have the necessary tools

[solid notch]

is location of root one of its topic, this is the question from that topic.

[worldly mesa]

And the numbers are ugly

[plucky rover]

Sometimes you need to get your hands dirty

[vague anchor]

Yes

[vague anchor]

Fr😭

[plucky rover]

Also I do not appreciate you calling the three people trying to help you "bad helpers"

[solid notch]

hence i suggested to mark and move on

[plucky rover]

No, solve it using what you have now and come back later

[solid notch]

yeah thats fair too

[vague anchor]

Umm i didn't mean that

[plucky rover]

Should still understand how to solve it on the base level

[solid notch]

it was implied

[plucky rover]

Anyway go do this

[vague anchor]

Oh universe lend me ur energy
I'm gonna solve this ugly

[plucky rover]

(I was hoping he'd make that realisation himself)

[solid notch]

yeah true fair mb

[vague anchor]

I was thinking but,.....

[vague anchor]

Were u IITian

[plucky rover]

Nah, I got the JEE trauma as inheritance

[vague anchor]

Btw how do ppl become helper here?

[plucky rover]

It's a role you self assign isn't it

[vague anchor]

Idk

[vague anchor]

Then ig ull get that soon

[solid notch]

hes talking about the green role

[plucky rover]

No I don't want to be pinged whenever people need help

Oh helpful, that idk

[vague anchor]

Btw I guess like how helpers spent their time here they might be paid but no way

[solid notch]

its auto assigned to the person who is actively helping ig

[plucky rover]

Ehhh no maths research leaves you with a lot of downtime

[solid notch]

nah most people here are highschool/ college math students , so yeah free time pass

[plucky rover]

At least if you manage your time properly

Anyway, stop talking and go get your hands dirty

[vague anchor]

That equation is literally becoming more than like 2 lines

[solid notch]

?

dont mindlessly multiply things out bro

[plucky rover]

Collect the coefficients

[vague anchor]

Btw why is e so used why is it a natural no. Like one that is recurring

[plucky rover]

The original discovery of e was when studying compound interest

[vague anchor]

Dirty equation

[plucky rover]

But the one you might appreciate the most is the limit formulation

[vague anchor]

Oh great
CI that i studied in 8th grade

[plucky rover]

e is the limit of (1+1/n)^n as n approaches infinity

[worldly mesa]

A lot of graduates

[solid notch]

higher levels of that

[vague anchor]

Ik

[solid notch]

yeah thats fair , **mostly **graduates are more active in the higher chanels or if a higher question is asked in the help chanel
which is lwk fair

[worldly mesa]

Yeah...

[plucky rover]

Unfortunately very few people here are asking what I'm working on so I'm stuck with JEE trauma

[thin cloud]

grinding help channel and you will have that role

[vague anchor]

How many hrs spent?😜

[solid notch]

bro do your question , talk in #discussion if you have smthing else

[vague anchor]

Sry

.close

[final saddle]

Channel closed

Closed by @vague anchor

Use .reopen if this was a mistake.

[vague anchor]

Ty

Guys

[vague anchor]

Xavier's specimen

[final saddle]

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.

[vague anchor]

6th one
I'm stuck in between

[final saddle]

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

[tired walrus]

do that, then i'll share my idea if applicable.

[vague anchor]

,rccw

[tired walrus]

your work looks kinda hard to read ngl

can you rewrite all this but calligraphic please

[vague anchor]

Lemme Redox it

[tame gust]

I find the problem statement really funny:

...then a value which is greater than |n+m| is
...I'd just choose the highest value out of spite >:D

[vague anchor]

It's multiple correct 😑

[tame gust]

ah, again it is weirdly written as if they are asking for one

[vague anchor]

I ain't a fool neither are they

Here it is Ann

[tired walrus]

ok let's see

so D ≥ 0 yep

but i think you set up the discriminant incorrectly

$\frac{3x^2+mx+n}{x^2+1}=y \to yx^2 + y = 3x^2 + mx + n \to (y-3)x^2 + mx + (n-y) = 0$, which gives $a=y-3$, $b=m$ and $c=n-y$.

[soft zealot]

Ann

[tired walrus]

so either you did some funny business in your head or you messed it up

or both! who knows.

but i would have expected $$m^2 - 4(y-3)(n-y) \geq 0$$

[soft zealot]

Ann

[vague anchor]

Ummm, sure?

[vague anchor]

It's just I solved it😓
No one understands me😭

[tired walrus]

well you may have solved it but then you didn't communicate it properly.

i for sure was confused at what you idd.

did*

also i just realized the x^2 coeff in my thing should be 3-y and not y-3, oops.

[tired walrus]

and "not communicated properly" qualifies as funny business, btw.

[vague anchor]

Yeah I was thinking the same
But I was not so courageous to oppose an post-graduate😅

[tired walrus]

oh come on

[vague anchor]

Well that doesn't matter

Let's go ahead

[tired walrus]

just because i have a master's degree does not make me a goddess who is never to be doubted or crossed

anyway, $$m^2 - 4(3-y)(n-y) \geq 0$$

[soft zealot]

Ann

[vague anchor]

Yepp next

[tired walrus]

the solution set of this ineq should be [-4, 3] but for now we should only keep that in the back of our mind.

and instead simplify this inequality. carefully, step by step, and without funny business.

[vague anchor]

Hdyk

[vague anchor]

I did and it is right

[tired walrus]

original question says the range of $\frac{3x^2+mx+n}{x^2+1}$ should be $[-4,3]$. the equation $\frac{3x^2+mx+n}{x^2+1}=y$ should have solutions for $x$ precisely when $y \in [-4,3]$ (no more, no less).

[soft zealot]

Ann

[tired walrus]

and what are your m and n

if you solved the question then why are you even here

[vague anchor]

U mean that I have to compare both equations?

[tired walrus]

... ok, we are definitely not understanding each other

wtf did you mean when you said "i did and it is right"

[vague anchor]

I meant that i simplified that equation 😅

[tired walrus]

inequality*

[vague anchor]

As u say

[tired walrus]

so whats this then, is that the simplified version?

[vague anchor]

Yepp

[tired walrus]

ok yeah i retraced your steps and you didnt fuck up

now you need this quadratic in y to be equal to 4(y+4)(y-3).

(for all y)

so yes you compare.

[vague anchor]

U r belief that i will f up is absolutely absolute

[tired walrus]

no?

i do not believe that you will fuck up

it is a "trust but verify" situation

[vague anchor]

Like Why?
I did these just following my intuition

[tired walrus]

my gripe was that your work was mostly unwritten/mental which means unverifiable to external parties (everyone except you)

[vague anchor]

Leave it

[tired walrus]

idrk how best to explain it other than "these both are quadratic inequalities so the one with m and n has to have roots exactly where the other one does"

[vague anchor]

Why whats the relation😭

Oh

Actually both are same

Both represents y

But why less than or equal to 0

I'm f up

[tired walrus]

one moment...

the quadratic 4y^2 - 4(n+3)y + 12n-m^2 has to look like this

[vague anchor]

Isn't it the range
Why would I get it as roots

[tired walrus]

ok look

the range is an interval, yes?

yes or no

[vague anchor]

Yes
It's the all values of y

[tired walrus]

you've written down an inequality whose solution set is that same interval

[vague anchor]

Why how?

[tired walrus]

ie you have passed from a "find the range of this function" exercise to a "solve this inequality" exercise...

[tired walrus]

cause of your discriminant ≥ 0 thing??

[vague anchor]

Oh yeah that should be greater than 0
Bcoz y does have roots

[tired walrus]

lemme try to illustrate the logic here

[vague anchor]

Sure 🙏

[solid notch]

your solution feels you are solving it like a set algorithm i dont see a thought process here

[tired walrus]

you've got 3 statements here

and the logic that underpins solving this question is the logical equivalences between these statements

this requires higher-level awareness than simply the ability to solve an inequality

[vague anchor]

Wait

[vague anchor]

I said that it was my intuition

[tired walrus]

vibe mathing

[vague anchor]

Yepp

[vague anchor]

Truly i was doing that

[tired walrus]

mmmmmmmmmmmmmmmmmmmmmmm no good.

[vague anchor]

😶‍🌫️

So now why we took (y+4)(y-3)<=0

[solid notch]

inequality for the y , given in the question

[vague anchor]

It's range why is it related with that

[plucky rover]

Because that's an inequality which is true exactly for that range

(do you see why)

[solid notch]

consider the function f(x)
if i write x>0 ,
can you tell me whats the domain

[tired walrus]

let me add to my diagram...

[plucky rover]

Okay did you understand the first part of what Ann did

[tired walrus]

there are 3 equivalences here which i've conveniently color-coded

[plucky rover]

(two of them look black to me)

[tired walrus]

"known from theory" is green.

[vague anchor]

Understood

[plucky rover]

Colour blindness kicking in

[tired walrus]

my aim is for mathemystic to:

• understand the red equivalence, AND recognize that it is question data
• understand the green equivalence, AND recognize that it is a manifestation of theory
• understand the black equivalence, AND recognize it follows from the other two, AND recognize that it is the path forward

[vague anchor]

My, my, thank u guys for doing this so much for me🥹

[vague anchor]

I think I've read the question well

[plucky rover]

Do you recognise any of those

[vague anchor]

Idk

I just know that u made me understand despite of colours
Ty

.close

[final saddle]

Channel closed

Closed by @vague anchor

Use .reopen if this was a mistake.

[vague anchor]

🥹

[final saddle]

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.

[robust horizon]

Problem Statement: Given an array and a sum k, we need to print the length of the longest subarray that sums to k. how do i start with this? i know i can check which elements equal to the sum given, but how exactly do i keep track of the longest sub-array?

[robust horizon]

one thing i can think of is:

i keep track of all the sub-arrays whose sum equal k

and then check which one has the maximum number of elements

but i think that would result in a lot of memory wastage

plus would not be convenient

[plucky rover]

Are you familiar with dynamic programming

[robust horizon]

no, not yet

[plucky rover]

That's gonna make this difficult lol

What have you done so far

[robust horizon]

i am not really sure how i am going to start with it, but i provided which direction my mind is going in

[plucky rover]

No I meant if this is for a course, what types of algortihms have you done so far

[robust horizon]

it is not for a course. i have just started learning arrays

i am not following any course

[plucky rover]

I see

If you haven't done dynamic programming, all you really have for this is brute force

[robust horizon]

not sure what exactly do you mean by that

[plucky rover]

Dynamic programming is a type of algorithm essentially

Until you learn that, I don't think you can solve this in any way that isn't to just check all subarrays

(which is what your approach was, I don't think you currently have the tools to do better)

[late rose]

well if you're not concerned about speed, you can get all subarrays, check which of those sum to k and then return the longest of those

[plucky rover]

That's already her approach

[robust horizon]

but would that not result in a lot of memory/cost wastage?

[plucky rover]

It would yes

But until you have some better tools it is the best you can do

Look up dynamic programming if you wanna explore that

[robust horizon]

what exactly does it come after? i really want to go step-by-step

[meager hedge]

I have a solution

[robust horizon]

yes please

wait

can you guide me to it? rather than giving me the whole thing

[plucky rover]

I'm not sure, I studied algorithms very out of order lol

[opal plinth]

That's quite the jump in difficulty

[robust horizon]

i want to build-up my logic thinking

[opal plinth]

@plucky rover I don't see how dynamic programming helps here

[plucky rover]

Wait hold on have I been saying dp instead of hashing

[opal plinth]

I don't see how hashing helps here

[plucky rover]

Hashing definitely helps here

[robust horizon]

i know a little about hashing, in theory

[meager hedge]

Ok here’s the thing, assume your array has n Numbers. Test all n arrays (There’s 1), then test all n-1 arrays (There’s n), etc.
Now think of the kind of test you wanna include

[opal plinth]

Is that really your solution?

[plucky rover]

That's still O(n²)

[meager hedge]

Yes, use a while loop and exit as soon as the Value drops beneath k

[plucky rover]

Wait Nel, I've been thinking contiguous subarrays

That's why I jumped to hashing (and called it DP)

[opal plinth]

I can't really imagine how you'd code that, so I can't tell what's wrong with your approach

There's a simple¹ O(n) solution for this problem
-# ¹when you've already solved this sort of problem before

[meager hedge]

It’a gonna be using nested loops

[plucky rover]

Using ||prefix sums|| or something different?

[opal plinth]

How is it better than OP's first thought?

[meager hedge]

I didn’t say it was

[opal plinth]

Fair enough

[plucky rover]

Cuz if so this is what I've been referring to as hashing, cuz you use a hashmap

[opal plinth]

Not really

[meager hedge]

It’s faster because you start by testing the longest sub arrays first

[plucky rover]

Oh hmm

[opal plinth]

For an array of 1000 elements, all 1s, with k=2, that's not faster

@robust horizon I can try to guide you through this but it is a difficult problem for someone who has just started using arrays...

[robust horizon]

can i try?

[opal plinth]

Of course

[late rose]

isnt it O(n^3)?

since summing is an O(n) operation

[robust horizon]

what would be the very first step i should be thinking of?

[plucky rover]

Shush all arithmetic operations are constant time for simplicity

[opal plinth]

No they're right it's O(n^3)

[plucky rover]

No I know I was being facetious

[opal plinth]

Maybe make an example array and try going through it manually, keeping track of relevant things, and see if you can find a solution that way

[robust horizon]

okay, say the array is {1, 2, 3, 5, 9}; with n = 5 and, say, k = 10

we know that there's two sub-arrays which equal to 10, i.e., {2, 3, 5}, and {1, 9}

oh

[opal plinth]

Oh, before we continue, I should ask: does the question define a subarray as contiguous? It should, but we can't be too sure...

[plucky rover]

(if it doesn't this is np-complete)

[robust horizon]

i copy-pasted the problem

[plucky rover]

I think subarray implies contiguous

vis a vis substring

[opal plinth]

It does

But clearly fudge doesn't quite know the terminology

[robust horizon]

i am sorry

[plucky rover]

Fair enough, also I just realized how to do this without hashing lol I was overcomplicating

[opal plinth]

It's normal, you're learning

[robust horizon]

so this would not work?

[opal plinth]

Alright so it's almost certainly contiguous subsequences

No it wouldn't

Let me try making an example

{3, 2, 1, 1, 1, 1, 3}, k=4

That should do the trick

[robust horizon]

question, what happens if it is not contiguous? i mean it makes it more complicated, yes, but aren't we just checking if elements of the array equal the sum? like how does it make a difference?

[opal plinth]

It makes a big difference in complexity, because there are 2^n subsequences (not necessarily contiguous)

[robust horizon]

okay, but what if we skip that part? would we change our approach?

[opal plinth]

and there is no known algorithm to solve this in non-exponential time IIRC

[plucky rover]

It's np-complete yes

[robust horizon]

np-complete?

[opal plinth]

Well in this example if we're considering subsequences, you'd need to consider things like {2, 3}, {3, 3}, {2, 1, 3}, and so on

[plucky rover]

Essentially it's a class of very hard problems that can't be done in polynomial time (as of now)

[robust horizon]

even if they do not equal the sum?

[robust horizon]

i see, thank you

[plucky rover]

You won't know until you check

So you have to check still

[opal plinth]

How do you know whether they do before you even consider them?

You know how, if you want to find the maximum value in an array, you need to check every single element?

[robust horizon]

how did you get {2, 3}, {3, 3}, {2, 1, 3} again?

[robust horizon]

yeah

[robust horizon]

like, are we not talking about contiguous sub-arrays?

[opal plinth]

Well in this problem you'd need to check every single one of the 2^n subsequences, and that's potentially a very big number

[robust horizon]

right

[opal plinth]

We are, I was just demonstrating the difference

[robust horizon]

so basically kind of like all the power sets?

[robust horizon]

oh okay, makes sense now

[opal plinth]

It's the power set, yes

But, going back to your problem, you only need to consider subarrays, and those are the contiguous kind

[robust horizon]

can i state what approach i am thinking of?

[opal plinth]

Go ahead

[robust horizon]

i am not sure if this is correct but

{3, 2, 1, 1, 1, 1, 3}, k=4

okay so

we start from the very first element, 3

we check if the next element plus itself equals 4

it does not

we keep an index?

so we know it is not {3} or {3, 2}

[robust horizon]

or is lesser than 4

[robust horizon]

so we move the index to 2

again, we repeat the process

3 is <= 4

so it could be {2, 1, ...}

[opal plinth]

You mean 2 <= 4?

[robust horizon]

oh, yes, sorry

[opal plinth]

Ok, so you get to {2, 1, 1}

[robust horizon]

then 2, 1

yeah

we get a sub-array

whose elements equal to 4

now we move our index to 1 (the first one)

and so on

but isn't that a super long process too?

wait

isn't that what the first thing i thought of, was

[opal plinth]

Kinda, but it's a start

I should've asked too, is it an array of positive integers only?

Do you have a link to the question?

[late rose]

yeah i was gonna say, negative numbers would throw a wrench in his solution

[robust horizon]

it was shared to me by a friend as a screenshot, is it okay if i forward the picture?

[opal plinth]

Yea

[opal plinth]

That formatting is wack but it does say "positives"

So I'm gonna take it as a yes

[robust horizon]

so, this is not exactly what we are trying to think of here

[opal plinth]

Well it's not bad, just not optimal

I think you should implement your solution before trying to improve on it

[robust horizon]

okay, let me work on it

would i need an inner loop for this?

[opal plinth]

Yes

[opal plinth]

If the problem included negatives in the array, you'd be right; I was already thinking of only positives, which has a (slightly) better solution

Also I found the source (spoilers because solutions): ||https://takeuforward.org/data-structure/longest-subarray-with-given-sum-k/||

[robust horizon]

question

so i would need to check for individual elements too, right

would that not result in a lot of ifs conditions?

[opal plinth]

Yeah, in the question's example there is a 5 in the array, and k=5

[opal plinth]

It shouldn't

[upper leaf]

||hashmap& prefixsum|| i think

[robust horizon]

can i work on it like how we work on the fibonacci series

[upper leaf]

like compute the nth fibonacci number?

[robust horizon]

no, how we calculate the sum

well that does not make sense

never mind

[upper leaf]

well for now what have you consider trying?

(after some help)

[robust horizon]

i am thinking

say i find a sub-array which equals k

where do i store it?

or do i just remember the indexes?

[opal plinth]

The question only asks for the maximum length of valid subarrays

[robust horizon]

but we wouldn't know the maximum length until we have found all such sub-arrays, right?

[opal plinth]

Of course, but you don't need to remember all subarrays, just the maximum length

[robust horizon]

ah

like the previous question

[opal plinth]

I don't quite remember but yeah probably

[upper leaf]

Problems like these challenges the optimality of your solution

You can brute force or use an effective algorithm and data structure or whatever else

[robust horizon]

um, something like this? #include <iostream> using namespace std; int main() { int a[7] = {3, 2, 1, 1, 1, 1, 3}, k = 4, sum = 0, max = 0; for (int i = 0; i < 7; i++) { for (int j = 0; j < 7; j++) { sum += a[j]; if (sum == k) { int l = j - i + 1; if (l > max) max = l; } } } }

[robust horizon]

honestly, i am just trying to think of how to implement it right now, no matter which way

[upper leaf]

well thats one way to do it but reset the sum

[opal plinth]

Did you mean to start j at 0?

[upper leaf]

start j at i

and cout the results

[robust horizon]

ah yes, i

[robust horizon]

after sum += a[j]?

[upper leaf]

well outside the loop

[robust horizon]

right, sorry

[opal plinth]

You need to reset sum at some point

I'll also give you a small improvement: you can start loop j from 0 to n-i; that immediately gives you the current subarray length

[robust horizon]

i do not follow how

[robust horizon]

after each ith iteration?

or no?

[opal plinth]

Yes

[robust horizon]

using namespace std;
int main()
{
    int a[7] = {3, 2, 1, 1, 1, 1, 3}, k = 4, sum = 0, max = 0;
    for (int i = 0; i < 7; i++)   
    {
        sum = 0;
        for (int j = i; j < 7; j++)  
        {
            sum += a[j];            
            if (sum == k)
            {
               int l = j - i + 1;
               if (l > max)
                max = l;
            }
        }
    }
    cout << "maximum length of the required subarray is: " << max;
}```

is okay now?

[opal plinth]

Should be

[upper leaf]

yeah

congrats

[robust horizon]

yay

[opal plinth]

Ok this is just details so I'll give you a version of your code that is functionally equivalent but reads better:

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int a[7] = {3, 2, 1, 1, 1, 1, 3}, k = 4, maxLength = 0;
    for (int i = 0; i < 7; i++)   
    {
        int sum = 0;
        for (int j = 0; j < 7-i; j++)  
        {
            sum += a[i+j];
            if (sum > k)
            { // optional, to not unnecessarily go to the end of the array
                break;
            }
            if (sum == k)
            {
               maxLength = max(maxLength, j+1);
            }
        }
    }
    cout << "maximum length of the required subarray is: " << maxLength;
}

Hopefully I didn't break it

[robust horizon]

i am trying to understand this: for (int j = 0; j < n-i; j++)

so

we start from 0

a[0]

[opal plinth]

I did break it, should've been sum += a[i+j]; (I edited just now)

[robust horizon]

also i did not declare n

so it should just be 7 - i

bad practice, sorry

[opal plinth]

True

[robust horizon]

i will fix it

so

[opal plinth]

You can also use size(a)

[robust horizon]

isn't it sizeof()?

or am i wrong

[opal plinth]

No, that would give you the number of bytes

int is more than one byte long

[robust horizon]

yes, makes sense

anyway

well

a five-minute break

head hurts

[opal plinth]

Sure

I'm gonna have to go, so if nobody answers, you can always just close and open a new channel

[upper leaf]

o7

[robust horizon]

okay, thank you so much for all the help!

for (int j = 0; j < 7-i; j++) does this not do something like this:

for i = 0, a[0] + a[1] + ... + a[6]
for i = 1, a[0] + a[1] + ... + a[5]

and so on?

[upper leaf]

Its supposed to be j < 7

[robust horizon]

right

sorry

fixed it

[upper leaf]

Does your code gives correct results now?

the next lune should be sum+= a[j]

line*

[robust horizon]

they did that before, but then changed it to sum += a[i + j]

[upper leaf]

Well why?

[robust horizon]

i also still do not get how it would be a better solution

[upper leaf]

Oh

[robust horizon]

because contiguous array?

or no?

[upper leaf]

Your solution and their given solution is kinda the same

[robust horizon]

yes, they mentioned it is just for better readability

[upper leaf]

Well yeah other than that theres not much improvement

[robust horizon]

okay so now that we have figured this one

what would be a better approach for this?

[upper leaf]

Well using hashmap and an algorithm called prefix sum

[robust horizon]

i know prefix sum but i do not see how exactly it helps

[upper leaf]

which will give you a solution for both negative numbers introduction and O(n) runtime

[robust horizon]

what exactly is the functionality of hashmap?

[upper leaf]

Its stores the prefix sums

[robust horizon]

😭

right

[upper leaf]

Basically instead of checking every subarray like your code

the optimal solution

uhm

It kinds of remembers the sums idk how to explain it

which let the code to not check alot of times and result in better runtime

[robust horizon]

i see, thank you

will close the channel now, thanks for the help!

.close

[final saddle]

Channel closed

Closed by @robust horizon

Use .reopen if this was a mistake.

[robust horizon]

.reopen

[final saddle]

✅

[robust horizon]

wait

why does this not work for negative numbers?

like why does it matter at all?

[upper leaf]

negative numbers doe work on your current code but it slows it down alot

[robust horizon]

how so?

[upper leaf]

Because then there will be no early termination and keeping track of the sum is hard since it can now decrease instead of increase

[upper leaf]

They added sum > k here to avoid neg nums

[robust horizon]

yes but does that not mean that if we do not add that condition, like i did not, earlier, it slows the process down for positive numbers too?

that is just, bad?

[upper leaf]

Your code works fine for postive integers

though there is a better approach but it works

[robust horizon]

why if we do not do sum > k

is that not the same thing

[upper leaf]

sum > k strictly prohibits negative numbers

[robust horizon]

yes, exactly

i am talking about

[robust horizon]

this

not nel's code

[upper leaf]

Well its both O(n²)

[robust horizon]

i do not get how

say our array has a thousand elements

[upper leaf]

Yes

[robust horizon]

it sure makes a big difference, no?

[upper leaf]

Yup

[robust horizon]

then why?

[robust horizon]

also how is this any different (better) than having negatives in our array

sorry if that is stupid

i am just trying to understand

[upper leaf]

both negative numbers and positive numbers results in the same runtime in your code

[robust horizon]

exactly!

that is what i was trying to confirm

[upper leaf]

Now you got it

[robust horizon]

yes, thank you

[upper leaf]

Sorry for the brief confusion lol

I keep thinking that the original questions asked for positive numbers only

[robust horizon]

no is okay, it was my bad

well, thank you still

.close

[final saddle]

Channel closed

Closed by @robust horizon

Use .reopen if this was a mistake.

[final saddle]

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.

[solar crest]

why is it that for this question even tho we're just rotating 1 funciton around some axis, we have to use the washers method?? why is it not the disk??

[solar crest]

i thoguht we're just rotating f(x) around the axis so i can just use f(x) - (-2) as the radius and then use the disk method

this is the solution, i dont rlly understand

[drowsy epoch]

because else you wouldn't had a hole

[solar crest]

?

wdym

[solar crest]

yea is it not like thar

[drowsy epoch]

You need to remove the -2 distance because else you wouldn't get that hole

[solar crest]

how can i know form the question that theres a hole..?

[drowsy epoch]

You are given a region

[solar crest]

doesnt it just say to rotate f(x) around the line y = -2?

[drowsy epoch]

,, R = {(x,y)\in\R^2 : (0 \le y \le 4-x) \land (0\le x \le 4) }

[solar crest]

OH

wait

[solar crest]

is this what they want?? the red part

i just saw it says between x axis

and f(x)

[drowsy epoch]

you are rotating the above part

yea

[solar crest]

OHHHHHHH

im tripping

tysm

but like

if it didbt say region between x axi

axis

and it just said to rotate f(x) around axis y = - 2

[solar crest]

then i can juist do this right?

[drowsy epoch]

most likely

saying rotate f(x) doesn't even make sense

you actually need to specify a region

[solar crest]

oh

😭

okay

ty

[drowsy epoch]

you would like

rotate a line

[solar crest]

oh

[final saddle]

@solar crest Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[final saddle]

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.

[urban wasp]

Solve $log(3y^2-10) to the base 4=2log(y-1)to the base 4+1/2$

[soft zealot]

Tan

[urban wasp]

so i used the power rule

$log(3y^2-10)to the base 4=log(y-1)^2 to the base 4 +1/2$

[soft zealot]

Tan

[urban wasp]

cancelled out log to the base 4

to get

$(3y^2-10)=y^2-2y+1) +1/2$

[soft zealot]

Tan

[umbral steppe]

$\log_{4}(3y^2-10) = 2\log_{4}(y-1) + \frac{1}{2}$
did you mean this?

[urban wasp]

dont know how to proceed

[tired walrus]

\log btw

[urban wasp]

yes

[soft zealot]

Lute

[tired walrus]

this was wrong btw

you tried to do 4^ both sides, right?

[urban wasp]

?

no i cancelled out log

[tired walrus]

you can't do that unless you've got JUST log(...) on the left and JUST log(...) on the right

[urban wasp]

i do tho

[tired walrus]

no you dont

you got that +1/2 on the right

[umbral steppe]

you still have the 1/2.

[tired walrus]

that thing doesnt just stay as-is yknow

[urban wasp]

oh

[tired walrus]

you need to simplify $2 \log_4(y-1) + \frac12$ into JUST $\log_4(????)$ (with \textbf{nothing} outside the log)

[soft zealot]

Ann

[urban wasp]

well i cant do anything abt it since if i subtract it from the rhs it appears on the lhs

[tired walrus]

you can write $\frac{1}{2}$ as $\log_4(4^{1/2})$

[soft zealot]

Ann

[urban wasp]

oh

and then rewrite the rhs as a single logarithm?

[tired walrus]

yes (which i told you to do to begin with)

personally i think this alternative is a bit clunkier than simply applying the function 4^x on both sides but like you do you

[urban wasp]

how do u apply 4^x

i dont think we've learned that

[tired walrus]

you go from $$\log_4(3y^2-10) = 2 \log_4(y-1) + \frac12$$ to $$4^{\log_4(3y^2-10)} = 4^{2 \log_4(y-1) + \frac12}$$ and simplify from there

[soft zealot]

Ann

[tired walrus]

yknow like, if a=b then 4^a = 4^b, you feel me?

[dusty quarry]

does it, though? (referring to OP's now deleted message)

[urban wasp]

yeah

[dusty quarry]

oh lmao rip message

[urban wasp]

wait

do 4^log cancel out?

$/4^/log{4}(3y^2-10)$

[soft zealot]

Tan

[urban wasp]

this might be the worst texit message ever generated

i need to learn how to use this thing man

[tired walrus]

yes that you do

[umbral steppe]

$b^{\log_b(x)} = x$

[soft zealot]

Lute

[tired walrus]

$4^{\log_4(3y^2-10)}$ does cancel out cleanly to just $3y^2-10$, yes. but on the RHS it is more interesting.

[soft zealot]

Ann

[urban wasp]

$4^/log*{4}(3y^2-10)*$

[soft zealot]

Tan

[dusty quarry]

what's with the forward slashes

[urban wasp]

well u just use the power rule and make it4^log(y-1)^2 and then cancel right?

[urban wasp]

oh

$4^\log*{4{(3y^2-10)*$

[soft zealot]

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

[dusty quarry]

and what's with the *s

[urban wasp]

im making it itallic

[dusty quarry]

you don't have to

[umbral steppe]

$\log_{4}(3y^2-10) = 2\log_{4}(y-1) + \frac{1}{2}$

those are underscores, and are used to write subscripts in latex.

[urban wasp]

what is a subscript

bases?

$4_2$

[soft zealot]

Tan

[urban wasp]

YES

[umbral steppe]

bases are not subscripts, but we write bases in subscripts, yes.

[urban wasp]

IM RIGHT FOR ONCE

[urban wasp]

why did it not appear in ur message tho

[umbral steppe]

because I did not escape the underscores, nor did I intend to.

[urban wasp]

can you elaborate

i dont understand

[umbral steppe]

?

[urban wasp]

wdym escape the underscores

[dusty quarry]

why did what not appear in Lute's message?

[urban wasp]

_

the underscore

[umbral steppe]

underscores are special characters in discord. you can surround text in one underscore (to either side) to italicize a phrase. surround a text with two to underline it.

[urban wasp]

wait in irl mathematics can we use _ to define subscripts like we use ^ to define exponents

[umbral steppe]

to show surrounding underscores as is, you need to escape them.

[umbral steppe]

no.

[urban wasp]

what

oh

okay

ty

but wait

[umbral steppe]

in real life you can write subscripts directly to begin with.

[urban wasp]

i wrote 4^1/2 as 2

but now

do i expand (y-1)^2 first?

[urban wasp]

yeah but people get confused right

thats why they invented the carrot

karat

Caret

[umbral steppe]

no. writing superscripts and subscripts properly makes people less confused than spamming _ and ^.

plus, the caret was not invented for this purpose, as far as I understand.

[urban wasp]

i see

can you help me tho

[urban wasp]

w this

[dusty quarry]

what is your current step

[urban wasp]

log(3y^2-10)=log(y-1)^2+log4^1/2

i wrote 4^1/2 as 2

(btw all the bases are 4)

and now

i want to write the rhs as a single log

do i expand (y-1)^2 first or just multiply to get (2y-2)^2 and then expand

[umbral steppe]

I don't see a need to do either. you can directly write the RHS as a single log from here.

[urban wasp]

yes

but how

[dusty quarry]

how do you normally simplify log x + log y if they're the same base

[urban wasp]

logxy

right?

[dusty quarry]

then do the same thing here.

[urban wasp]

2(y-1)^2?

or

y^2-2y+1(2)

oh okay

i got

3y^2-10=4y^2-8y+4

(after cancelling log)

0=y^2-8y+14??

[dusty quarry]

keep going

[urban wasp]

factors of 14:1,2,7,14

therefore

this question has no solutions

wait no the stupid formula

uhh

,w calculate 64-4(14)

[soft zealot]

Results provided by WolframAlpha

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

Requested by tan211111

[urban wasp]

i hope wolfram used pemdas

okay

(8+-8)/2

is my final answer

[bold turtle]

Sorry, how did you get this line?

[urban wasp]

cancel lol

log*

[bold turtle]

No, I mean

The last message you sent where like three of us responded , that's correct

How did you get from that, to this?

[bold turtle]

Like, the RHS after cancelling logs became this

Did you misunderstand this to mean you square the 2 as well? (which is incorrect)

[urban wasp]

what

[urban wasp]

after cancelling log, it became this.

[bold turtle]

No, after cancelling logs it became " 3y^2 - 10 = 2(y-1)^2 "

Might I ask for a

!xy

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[umbral steppe]

I think we can proceed from here and explain why OP may be mistaken.

[final saddle]

@urban wasp Has your question been resolved?

[urban wasp]

this

@bold turtle

[bold turtle]

I mean, show me the source

If it's from a book, take a photo

If it's online, take a screenshot

[urban wasp]

there is nothing more.

its just

Solve $\log_{4}(3y^2-10)=2\log_{4}(y-1) + \frac{1}{2}$

[soft zealot]

Tan

[urban wasp]

This

[bold turtle]

Okay, so do you agree with getting to this line?

[urban wasp]

no

[bold turtle]

$3y^2 - 10 = 2(y-1)^2$

[soft zealot]

Waes (Wires)

[urban wasp]

cuz i heard u cant cancel logs unless the coefficient is 1

[urban wasp]

oh this

yes

[bold turtle]

Okay

So you've mis-expanded the RHS then

How would you expand (y-1)^2?

[urban wasp]

y^2-2y+1

[bold turtle]

Right

So now we have two lots of (y-1)^2

i.e. 2 (y-1)^2

So what's this when expanded?

[urban wasp]

=(2y-2)^2
=4y^2-8y+4

[bold turtle]

No

[urban wasp]

or do u expand first

and them multiply by 2

[bold turtle]

You've got a multiplication and a power

Which comes first?

[urban wasp]

2y^2-4y+2

[bold turtle]

There we go

[urban wasp]

parenthesis

[bold turtle]

Okay, so are we going to take those parentheses and square first, or multiply by 2 first?

[urban wasp]

bro

arent they the same thing???

simplify them

[bold turtle]

?

[bold turtle]

This...

[bold turtle]

...is clearly not this

[urban wasp]

divide by 4

y^2-2y+1

[urban wasp]

divide this by 2

y^2-2y+1

[bold turtle]

Okay but that logic is worse than shit I'm gonna be fully honest

Because by the same token, "4" is the same as "8" because half of 4 is 2 and a quarter of 8 is 2

[urban wasp]

so is this a super ultra rare case where it just so happens to be that they have the commutative property?

[bold turtle]

?

[urban wasp]

and it wouldnt happen with any other(or few) equations

[bold turtle]

No this is more of an ultra rare case where your algebra needs more revision before continuing with logs ngl

2x^2 = 2 times whatever x^2 is; it is not equal to the square of (2x)

[urban wasp]

bro i even need help with addition and subtraction

sometimes i subtract things from the rhs and accidentally add them on the lhs

but yeah

thanks

[bold turtle]

Honestly at this point that's something you yourself need to practise

A person typing on a screen can only go so far to explain those concepts

[urban wasp]

i practice 3 hours a day

[bold turtle]

Practise what

I'm talking about basic algebra

[urban wasp]

addition, subtraction,multiplication, and division

[bold turtle]

...That's arithmetic

[urban wasp]

oh

[bold turtle]

Not algebra

[urban wasp]

isnt arithmetic a sequence of stuff

(n-1)(n-2)

typa stuff

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[bold turtle]

.reopen

[final saddle]

✅

[bold turtle]

You're thinking of arithmetic sequences (which you haven't described properly)

Those are called arithmetic because you're adding/subtracting between consecutive terms

[urban wasp]

so i suck at arithmetic

[bold turtle]

Similarly "geometric" sequences are so called because of having common ratios between consecutive terms

[urban wasp]

no wonder i always get polynomial questions wrong all the time

[bold turtle]

Arithmetic, the gerneral term, just means addition, subtraction, multiplication, division etc.

i.e., doing numerical calculations

[urban wasp]

i see

[worldly mesa]

arithemtic is usually just addition

[urban wasp]

calculators like wolfram can do more than arithmetic right

[worldly mesa]

yeah they can do a lot

[urban wasp]

,w differentiate 3x^2

[soft zealot]

Results provided by WolframAlpha

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

Requested by tan211111

[bold turtle]

I'm talking about elementary school arithmetic, i.e. numeracy

That context is important

[worldly mesa]

yeah

[urban wasp]

,w integral 3x^2

[soft zealot]

Results provided by WolframAlpha

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

Requested by tan211111

[bold turtle]

(and you'd missed the entire context for it lol)

[urban wasp]

im in hs..

[late rose]

who needs arithmetic skills when you have a calculator /s

[bold turtle]

ik, else you wouldn't be doing logs

But what I'm talking about is something that should be taught in elementary/middle school

That being said, I'm guessing you're from the US?

[urban wasp]

nah we havr to do 2 papers one without calc and one with calc so yeah

[late rose]

in high school??

[urban wasp]

yeah

[bold turtle]

i.e. the land of no regulations on their maths syllabus so fuxk me if I ever understand what's on their papers

[urban wasp]

for cambridge

[late rose]

oh dude i feel bad for you

[urban wasp]

bro

[late rose]

without calculator i literally cannot function

[urban wasp]

i have to write 9 PAPERS for chemsitry,biology,and physics

🙏

3 papers per subject

[late rose]

i mean

i also have 3 papers

[urban wasp]

worth like 180/320 marks each

[late rose]

one mcq, one open-ended, one practical

[urban wasp]

i got

okay

nvm

we are getting derailed

[late rose]

fr

have you solved the log question yet?

[urban wasp]

nope

but guess what

ik someone who can

,w solve $\log_{4}(3y^2-10)=2\log_{4}(y-1) + \frac{1}{2}$

nevermind

[bold turtle]

That shit only flies if you can acc type the question in properly

,w solve $\log_{4}(3y^2-10)=2\log_{4}(y-1) + \frac{1}{2}$

[soft zealot]

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[bold turtle]

wait a fokken second

[dusty quarry]

just replace all the y with x

[bold turtle]

,w solve for y $\log{4}(3y^2-10)=2\log{4}(y-1) + \frac{1}{2}$

[soft zealot]

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[dusty quarry]

WA loves its exes

[bold turtle]

It's got so many exes

[dusty quarry]

hang on

[urban wasp]

bruh

thought this calc was supposed to be smart

[late rose]

did you mean $\log_4(3y^2 - 10) = 2\log_4(y - 1) + \frac{1}{2}$?

[urban wasp]

cant even do basic exponential equations

(just like me fr)

[soft zealot]

0

[bold turtle]

Humans are idiots; computers are just really fast idiots

[urban wasp]

yes

[dusty quarry]

,w solve for y, log_4 (3y^2 - 10) = 2log_4 (y-1) + 1/2

[soft zealot]

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[bold turtle]

There we go

[urban wasp]

https://tenor.com/view/fnb-get-this-man-a-true-gif-25257480

[bold turtle]

Incidentally that's what I got by hand as well

[dusty quarry]

i should have checked your arithmetic

[late rose]

combine the expression on the right into one log

[dusty quarry]

that ship has sailed

[bold turtle]

So, 0, we've already gone through that

[dusty quarry]

anyway, OP, do you agree with WA's answer now

[urban wasp]

https://tenor.com/view/wtf-gif-5426180885264703865

[bold turtle]

We've gone as far as $3y^2 - 10 = 2y^2 - 4y + 2$

[soft zealot]

Waes (Wires)

[urban wasp]

ohhh

yeah that

i did it alr

just need help with basic algebra and arithmetic calculations

[dusty quarry]

that, we shall see you in #math-discussion ir #study-discussion unless you have specific questions

cuz help channels are not the place for a crash course in topics

but btw OP, just curious. are you above 15 years old?

[urban wasp]

remember when you taught me the laws of square roots

[urban wasp]

nope

[dusty quarry]

who

[urban wasp]

you

[dusty quarry]

idqr

i know another helpee fucked with me using square roots

it wasn't you

[urban wasp]

oh

[dusty quarry]

i see. i guess you're a peer then

[urban wasp]

hollup lemme try find it

[urban wasp]

what?

[dusty quarry]

what

[bold turtle]

,w peer

oh ffs

,w define peer

[soft zealot]

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[bold turtle]

Sense 1

[dusty quarry]

not exactly sense 1

[bold turtle]

Well, ... okay yh

[dusty quarry]

peer here = same age

[urban wasp]

i got it

[dusty quarry]

oh ok, cool

[urban wasp]

WHA

WOLFRAM IS A DICTIONARY TOO?!

,w give me 3 synonyms for nice

[soft zealot]

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[late rose]

WHAT