Math problem (high iq math nerds only)

[mogger123]

Saw this on twitter and can't see an easy way to solve it

If anyone solves it and explains their solution they will earn my respect

 

[Xangsane]

View attachment 2064250

Saw this on twitter and can't see an easy way to solve it

If anyone solves it and explains their solution they will earn my respect

Do you have a uni exam?

 

[mogger123]

Do you have a uni exam?

graduated ages ago

 

[Deleted member 4362]

-1

 

[Deleted member 4362]

2

 

[Deleted member 4362]

It’s not very difficult tho, but average iQ here is probably around 60 so I guess it’s relative

 

[Incellectually_Shy]

For someone with basic verbal ability, it is easy to solve using simple knowledge.

 

[Bashan]

For someone with basic verbal ability, it is easy to solve using simple knowledge.

So solve it motherfucker

 

[Deleted member 4362]

So solve it motherfucker

This @Incellectually_Shy muh verbal skills

 

[Deleted member 4362]

View attachment 2064250

Saw this on twitter and can't see an easy way to solve it

If anyone solves it and explains their solution they will earn my respect

Is that second one supposed to be f(x)^2/f(x)

 

[Deleted member 4362]

Cos sqrt[e^(fx) - x] is not an element of real if x goes to infinity

The 1st one you just rearrange + converge

I thought the second one is the same but I didn’t realise it’s f(x^2) not f(x)^2

 

[Deleted member 4362]

Idk I feel pretty dumb now but maybe it’s the inverse of the exponential then, use the natural log to get infinity. So the answer will be zero

 

[Mouthbreath]

View attachment 2064250

Saw this on twitter and can't see an easy way to solve it

If anyone solves it and explains their solution they will earn my respect

I call y = f(x)
Rewrite the first equation to e^y - y^2 = x => if y ->oo, then x has to go to infinity as well because x is roughly e^y for large y. In fact lim_(y -> oo) x ( y ) = e^y

So the the first question is lim_(y->oo) y^2 /e^y -> 0
Second question: lim_(x->oo) log(x^2)/log(x) =lim_(x->oo) 2*log(x)/log(x) = 2

 

[Deleted member 22627]

 

[Deleted member 4362]

I call y = f(x)
Rewrite the first equation to e^y - y^2 = x => if y ->oo, then x has to go to infinity as well because x is roughly e^y for large y. In fact lim_(y -> oo) x ( y ) = e^y

So the the first question is lim_(y->oo) y^2 /e^y -> 0
Second question: lim_(x->oo) log(x^2)/log(x) =lim_(x->oo) 2*log(x)/log(x) = 2

Tf nigga

1/infinity = 0

 

[Deleted member 4362]

I call y = f(x)
Rewrite the first equation to e^y - y^2 = x => if y ->oo, then x has to go to infinity as well because x is roughly e^y for large y. In fact lim_(y -> oo) x ( y ) = e^y

So the the first question is lim_(y->oo) y^2 /e^y -> 0
Second question: lim_(x->oo) log(x^2)/log(x) =lim_(x->oo) 2*log(x)/log(x) = 2

The limit is when x tends to infinity not the function of x.

You can’t approximate a function like that, what if it was tan(x) it would go to infinity at pi/2 & 3pi/2 & 5pi/2 & 7pi/2 …. u get the point

 

[Deleted member 4362]

@OldRooster what is the answer

 

[Mouthbreath]

The limit is when x tends to infinity not the function of x.

You can’t approximate a function like that, what if it was tan(x) it would go to infinity at pi/2 & 3pi/2 & 5pi/2 & 7pi/2 …. u get the point

x tends to infinity exactly when y tends to infinity as well.

e^y - y^2 = x

This function looks like x = e^y for large y, so it's just an exponential in that regime, so you can "invert" it. The initial function happens to be bijective as well. If it wasnt injective then the the reformulations that I wrote are not allowed (because you cant invert the function), you would need to be a bit more careful, but the end result would still be correct.

 

[Deleted member 4362]

x tends to infinity exactly when y tends to infinity as well.

e^y - y^2 = x

This function looks like x = e^y for large y, so it's just an exponential in that regime, so you can "invert" it. The initial function happens to be bijective as well. If it wasnt injective then the the reformulations that I wrote are not allowed (because you cant invert the function), you would need to be a bit more careful, but the end result would still be correct.

I thought the 1st question is y^2/x? why did you not use y^2/(e^y-y^2) ?

Also can u explain what u did in the 2nd with the logs, I haven’t used those in years

 

[Deleted member 4362]

I call y = f(x)
Rewrite the first equation to e^y - y^2 = x => if y ->oo, then x has to go to infinity as well because x is roughly e^y for large y. In fact lim_(y -> oo) x ( y ) = e^y

So the the first question is lim_(y->oo) y^2 /e^y -> 0
Second question: lim_(x->oo) log(x^2)/log(x) =lim_(x->oo) 2*log(x)/log(x) = 2

I don’t get it what did I do wrong the maths checks out, how we have different answers?

 

[Deleted member 4362]

I call y = f(x)
Rewrite the first equation to e^y - y^2 = x => if y ->oo, then x has to go to infinity as well because x is roughly e^y for large y. In fact lim_(y -> oo) x ( y ) = e^y

So the the first question is lim_(y->oo) y^2 /e^y -> 0
Second question: lim_(x->oo) log(x^2)/log(x) =lim_(x->oo) 2*log(x)/log(x) = 2

Actually nvm we have the same method for the second one. I didn’t expand the log in my working, it should have been 2*[f(x)/f(x)] which would equal 2.

But still don’t understand what u did in the 1st

 

[mogger123]

none of you have considered that f(x) could be any function so e^f(x) could grow at a smaller rate than just x on it's own

idk how to fix this problem without solving for f(x) directly, which doesn;t seem possible

maybe the answer wants separate cases or there is a way to prove this cant be the case

 

[lordgandy2000]

Not sure what retards are saying here but I wrote everything out with justification so u don’t have to look at retarded text formatting

Answer is 0 and 2 respectively

 

[mogger123]

View attachment 2064528
Not sure what retards are saying here but I wrote everything out with justification so u don’t have to look at retarded text formatting

wait nvm its squared

 

[Mouthbreath]

none of you have considered that f(x) could be any function so e^f(x) could grow at a smaller rate than just x on it's own

idk how to fix this problem without solving for f(x) directly, which doesn;t seem possible

maybe the answer wants separate cases or there is a way to prove this cant be the case

It can't be any function. The function is completely determined by the equation.

If you're so concerned about that. You can prove injectivity and surjectivity => Bijectivity and therefore invertibility

Why do you need separate cases?

 

[lordgandy2000]

wait nvm its squared

I’m quite sure my answer is correct. The only thing I skimmed over is why e^f(x) goes to infinity faster than f(x)^2 for the first limit.

 

[mogger123]

It can't be any function. The function is completely determined by the equation.

If you're so concerned about that. You can prove injectivity and surjectivity => Bijectivity and therefore invertibility

Why do you need separate cases?

but then how can you determine it's behaviour when the equation is unsolvable

I think lordgandy figured it out. you need to use that the square is always positive

 

[mogger123]

View attachment 2064528
Not sure what retards are saying here but I wrote everything out with justification so u don’t have to look at retarded text formatting

Answer is 0 and 2 respectively

can you explain the last line

 

[Deleted member 4362]

View attachment 2064528
Not sure what retards are saying here but I wrote everything out with justification so u don’t have to look at retarded text formatting

Answer is 0 and 2 respectively

Why is not 1? Why did my method not work

 

[Deleted member 4362]

none of you have considered that f(x) could be any function so e^f(x) could grow at a smaller rate than just x on it's own

idk how to fix this problem without solving for f(x) directly, which doesn;t seem possible

maybe the answer wants separate cases or there is a way to prove this cant be the case

Ova 4 our lowIQ

 

[lordgandy2000]

can you explain the last line

Sure! As I wrote in the 2nd to last line, the limit as x goes to infinity of e^f(x)/x = 1 by rewriting the terms from part (a)

So e^f(x) and x are equal as x goes to infinity.

Which means f(x)= ln(x) as x goes to infinity.

So f(x^2) = ln(x^2)

And by using logarithm rules, ln(x^2) = 2ln(x)

 

[lordgandy2000]

Why is not 1? Why did my method not work

You assumed incorrectly that e^f(x)/x tends to 0 as x tends to infinity.

This is not true, you can see from my answer that e^f(x)/x actually tends to 1 as x tends to infinity.

 

[jojoyou]

View attachment 2064528
Not sure what retards are saying here but I wrote everything out with justification so u don’t have to look at retarded text formatting

Answer is 0 and 2 respectively

This

 

[mogger123]

Sure! As I wrote in the 2nd to last line, the limit as x goes to infinity of e^f(x)/x = 1 by rewriting the terms from part (a)

So e^f(x) and x are equal as x goes to infinity.

Which means f(x)= ln(x) as x goes to infinity.

So f(x^2) = ln(x^2)

And by using logarithm rules, ln(x^2) = 2ln(x)

nice

well done, you win!

 

[lordgandy2000]

nice

well done, you win!

FYI @Mouthbreath had the correct answer too, he just expressed it in a weird way with the invertibility + typing math symbols out in text + skimmed over alot in his explanation

 

[Eduardo DOV]

View attachment 2064528
Not sure what retards are saying here but I wrote everything out with justification so u don’t have to look at retarded text formatting

Answer is 0 and 2 respectively

Couldn't you have used L'hospital instead of saying that one function grows much faster than the other?

 

[Eduardo DOV]

I studied this a long time ago, I don't remember very well

 

[lordgandy2000]

Couldn't you have used L'hospital instead of saying that one function grows much faster than the other?

Yeah in fact, that's the reason. e^f(x) / f(x)^2 is in indeterminate form so it applies.

The only thing is I'm assuming f(x) is differentiable on the open interval, which I'm just gonna assume as a technicality

 

[Deleted member 4362]

Yeah in fact, that's the reason. e^f(x) / f(x)^2 is in indeterminate form so it applies.

The only thing is I'm assuming f(x) is differentiable on the open interval, which I'm just gonna assume as a technicality

Are u a maths student?

Especially @Mouthbreath ? I study Elctrical Eng. so I was surprised you mentioned bijection etc. cos we use those to describe signals, I thought u were just some schizos.

it’s been a while since I did pure maths + I know it’s cringe cos a lot of people throw it around on this forum underservedly but u guy actually do seem highiQ.

What r ur career plans?

 

[Mouthbreath]

Are u a maths student?

Especially @Mouthbreath ? I study Elctrical Eng. so I was surprised you mentioned bijection etc. cos we use those to describe signals, I thought u were just some schizos.

it’s been a while since I did pure maths + I know it’s cringe cos a lot of people throw it around on this forum underservedly but u guy actually do seem highiQ.

What r ur career plans?

I studied physics.

Btw if you're doing electrical engineering, you might be interested in these videos I used to watch a long time ago:



Imo they're some of the best videos I have ever watched. You might already know this stuff, but even if I knew everything I would honestly still watch it, because it is so satisfying to see how a whole programmable computer works step by step and how it is made up of thousands of individual transistors. A modern cpu is obviously way more complicated, but a lot of the key concepts still apply.

 

[lordgandy2000]

Are u a maths student?

Especially @Mouthbreath ? I study Elctrical Eng. so I was surprised you mentioned bijection etc. cos we use those to describe signals, I thought u were just some schizos.

it’s been a while since I did pure maths + I know it’s cringe cos a lot of people throw it around on this forum underservedly but u guy actually do seem highiQ.

What r ur career plans?

I studied pure math. Actually has quite little to do with numbers and calculation per se, its more about logic and proofs.

Also probably got brain damage from it which is why I'm 26 yo KHHV and can't talk to women JFL

 

[Deleted member 22627]

It’s simple really, as x approaches infinity my dick expands exponentially in your bussy.

 

[RichmondBread]

Math is for sissies

 

[Deleted member 4362]

I studied pure math. Actually has quite little to do with numbers and calculation per se, its more about logic and proofs.

Also probably got brain damage from it which is why I'm 26 yo KHHV and can't talk to women JFL

Bro I built a robot last year + I’m 24 virgin in less than a month. Sexhaving is for monkeys.

Mirin pure maths + physics majors tho, there isn’t a lot of degrees i have respect for given what I do but I could never do pure maths or physics, mogs me hard.

What career have u perused?

 

[lordgandy2000]

Bro I built a robot last year + I’m 24 virgin in less than a month. Sexhaving is for monkeys.

Mirin pure maths + physics majors tho, there isn’t a lot of degrees i have respect for given what I do but I could never do pure maths or physics, mogs me hard.

What career have u perused?

Private equity

Tbh I might retire soon and just LDAR on .org and incels.is full time, I don't spend money at all except for food and internet so no point moneymaxxing

 

[Deleted member 21467]

holy fuck this thread is brutal

absolutely mogs me

 

[Deleted member 4362]

Private equity

Tbh I might retire soon and just LDAR on .org and incels.is full time, I don't spend money at all except for food and internet so no point moneymaxxing

Stats? Age, race, height, psl?

 

[lordgandy2000]

Stats? Age, race, height, psl?

26, hapa, 192cm, PSL probably lower but SMV higher because I look like some kpop twink

 

[Deleted member 4362]

26, hapa, 192cm, PSL probably lower but SMV higher because I look like some kpop twink

Lift fraud to 195 + gymmax you will have god tier smv. It’s Newtons law, you see a truck coming towards you what are you gonna do? Move tf out of the way.

I’m 188 barefoot + frauding 190-193 + I can see how ova it is for Manlets jfl

 

[Artemis]

Cos sqrt[e^(fx) - x] is not an element of real if x goes to infinity

The 1st one you just rearrange + converge View attachment 2064441

I thought the second one is the same but I didn’t realise it’s f(x^2) not f(x)^2

e^(f(x)) / x can be tending to infinity if f(x) is a function growing faster than f(x) = log(x).