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imqwerty
 one year ago
emergency!!!
imqwerty
 one year ago
emergency!!!

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Last one x = y = z = 4

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2any question that u like...these r the question which i wsn't able to solve

Empty
 one year ago
Best ResponseYou've already chosen the best response.8First one probably some fermat's little theorem nonsense, the second one looks like part of the Riemann zeta function so you could probably just multiply two geometric series or something together for it, and 3rd one looks like it's probably a telescoping series, just my first guesses

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2@ParthKohli how did u get x=y=z=4

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3It's a symmetric situation. Apply all inequalities you please, but you'll finally get to this.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Simplest way to think of it (not a proof) is that \[n^2 > (na)(n+a)\] when a is not equal to zero, so that's the max when they're all the same.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1first q doesnt look easy

Empty
 one year ago
Best ResponseYou've already chosen the best response.8I think we can split that 3rd one up using partial fractions, then it becomes a telescoping series that way.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Good questions by the way!

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2thanks @Empty @ParthKohli i can't get to the result x=y=z. can u tell how to start ..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I've only given a qualitative argument here so far  the whole situation you have here is symmetric in nature. x can be interchanged with y, y with z or z with x, and the situation won't change. So if the triplet \((a, b , c)\) gives you a maximum value then so should \((b, a, c)\) or \((b, c, a)\) or \((c, a, b)\), which means that \(x = y = z \) when the given expression is maximum. Let me think of an inequality that can be applied here. You'll see that we'll find the maximum when the equality of two expressions is considered, which is almost always when all variables are equal.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Ok so for the second one I'm getting: $$\frac{1}{19}\frac{1}{14}\frac{5}{6}$$ Does that match your answer?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2wait m doing the 1st one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1for first one you may want to try factoring it http://www.wolframalpha.com/input/?i=factor+5%5E5%5E%28n%2B1%29%2B5%5E5%5En%2B1

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Yeah I think it is more clear how to solve the first one if you do that, here to make factoring easier substitute: $$5^{5^n}=a$$ then you end up with the expression: $$a^5+a+1$$ which somehow wolfram is able to factor out into: $$(a^2+a+1)(a^3a^2+1)$$

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2@ParthKohli if we are given a question like  x+y=4 and x belongs to positive integers. and we are supposed to find out maximum of xy + x/y +y/x then by symmetric approach x=y=2 but in this question x=3 and y=1 gives the maximum so the symmetric approach is violated.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2@Empty we can't write 5^5^(n+1) as a^5 if a=5^5^n

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3OK, I didn't say that x = y gives you the maximum. I said it gives you the maximum if it exists. Here, x = y is most probably giving you the minimum.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2yes so how do we know that in which case x=y gives the max

Empty
 one year ago
Best ResponseYou've already chosen the best response.8$$5^{5^{n+1}}=5^{5*5^n} = (5^{5^n})^5$$ @imqwerty

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3You have to check if x = y is giving you the maximum/minimum by entering another pair of values. If that pair gives you a higher value, then x = y gives you the minimum.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i feel #1 is still a stupid q, you're not gona learn anything frm it

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2wait sry yes we can put a^5...

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2yes it is @ganeshie8 :D

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Y'know, qwerty, it's just like that with single variable functions. You can differentiate a function and equate it to zero, but you still have to check if that gives you the maxima or minima by other methods.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Ok so I also found a good way of solving the second one: We only want odd total powers, so the way we can do that is only combine even powers with odd and odd with evens. So in order to do that, we multiply the geometric series' together: $$\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) $$ factor out a 1/3 and 1/2 so you get only geometric series in terms of 1/4 and 1/9, and you're done!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Isn't that the standard way to do it though? How did you originally do it?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2the 1st one is solved thanks guys :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align} x^5 + x + 1 & = (x^5x^2)+(x^2+x+1)\\ \\ & = x^2(x^3  1) + (x^2 + x + 1) \\ \\ & = x^2(x1)\color{blue}{\bf (x^2 + x + 1)} + \color{blue}{\bf (x^2 + x + 1)}\end{align}\] http://math.stackexchange.com/questions/477295/factorx5x1

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3To be fair, that solution was most likely obtained by working backwards. :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that is okay for a proof

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Yes of course, but you can't really think of that on your own.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that all goes as "scratch work" :P

Empty
 one year ago
Best ResponseYou've already chosen the best response.8ahhh thanks @ganeshie8 that's a clever trick I think I can even remember that as adding fancy form of zero and then abusing the geometric series yet again lol

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I'm ashamed that I have no other ideas other than telescoping for #3.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8It wants to telescope, why force it to be anything it's probably not? :P

Empty
 one year ago
Best ResponseYou've already chosen the best response.8When you separate it out, the coefficients are both 1, it's truly a beautiful result

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I'm either dumb or too much drunk to notice it but why do you think telescoping is a bad idea ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Yeah you're not drunk enough ganeshie, cause this baby wants to see the moon

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I didn't say telescoping is a bad idea. I just said that this series doesn't seem to be telescoping nicely and I have no other ideas.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2m really confused nd annoyed nd kinda drunk ..cheers @ganeshie8 @Empty ..can u please tell me what u did with the 2nd question

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2my mind is jst nt wrkin normal today :(

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3OK, it's actually telescoping really nicely.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Here I made it smaller, part of the total sum, just write out all these terms (it's not tooo many, and it's good for you) \[\left(\sum_{i=0}^{3} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{3} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{3} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{3} \frac{1}{2^{2j+1}} \right) \]

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2no i mean hw did u split that

Empty
 one year ago
Best ResponseYou've already chosen the best response.8so by multiplying it out you can kinda see how I came up with this too... I'll draw a picture of how I think about it

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{6^k}{(3^k  2^k)(3^{k+1 }  2^{k+1})}\]\[= \frac{3^{k+1}}{2^{k+1}  3^{k+1}}  \frac{3^k}{3^k  2^k}\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.8dw:1438453754914:dw So this is the first half of the terms! @imqwerty

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Supposed to be\[= \frac{3^{k+1}}{2^{k+1}  3^{k+1}}  \frac{3^k}{2^k  3^k}\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Specifically I knew to write that because the only way for i+j=odd is when one is even and the other is odd since odd+odd=even and even+even=even. So combine all the 2s with even powers with all the 3s with odd powers, and add it to all the 2s with odd powers with all the 3s with even powers.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8so that's where the 2i and 2i+1 is coming from, keeping the powers even and odd: \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) \]

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2thank u so much @Empty :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3OK, so we're sorta done with all of the questions.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1whats the ansur for q2

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Are you asking me or asking imqwerty?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2m calculating the answer...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3@Empty Did you check out the solution to the problem I gave you yesterday?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Oh no I didn't, also I ended up getting 2 to that, I'll go check real fast

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2:) a big THANKS TO ALL OF YOU @Empty @ParthKohli @ganeshie8 :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Ganeshie and Qwerty haven't done that problem so I'll ask them to do it too.\[\sum_{n=1}^{\infty}\frac{F_n}{2^n}\]where \(F_n\) is the \(n\)th Fibonacci number. \(F_1 = F_2 = 1 \).

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2my answer the #2 is 121/144

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I am familiar with that problem \[\sum_{n=1}^{\infty}F_nx^n= \frac{x}{1xx^2}\] plugin \(x=1/2\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Here's the solution: \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) \] \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \frac{1}{3}\left(\sum_{j=0}^{\infty} \frac{1}{3^{2j}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \frac{1}{2} \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j}} \right) \] \[\left( \frac{1}{3} + \frac{1}{2} \right) \left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j}} \right) \] \[\left( \frac{1}{3} + \frac{1}{2} \right) \left( \frac{1}{13^2} \right) \left(\frac{1}{12^2} \right) \] \[\frac{5}{144}\]

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2damn my mind is nt ok today ....instead of starting with i=0 i started with i=1. :( feeling bad at the silly mistakes m doin :(

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Haha nah it's fine that's why I wrote it all out for you. And hey, if you want you can always come back tomorrow and look, no reason to feel bad XD

Empty
 one year ago
Best ResponseYou've already chosen the best response.8I'm pretty sure "feeling bad" is not supposed to be a step in solving problems haha, even though sometimes it does tend to be when those problems are proofs in real analysis for me lolol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1sometimes i hate problems from indian competitive exams/tests mostly because they are meaning less, how do they expect a 10/12th grader to know about generating functions and stuff ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1they just want the exams to be tough, they dont bother to look at the syllabus of 12th grader

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2the reason y m feeling bad and annoyed  today the results of the last test held at of coaching institute came the 1st ranker got 360/360 nd i ws like O_o nd i ws 12th with score 329. i could've done way better if i wuld have rechecked my paper after completing it instead of looking at other ppl faces :(

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2at my home m nicknamed  'silly mistake'

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1trust me, you and PK are way better than me when i was in my 12th

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Yeah I don't understand the obsession with rote memorizing all these useless onetimeuse tricks for problems mankind invented calculators and wolfram alpha for. Time to change the curriculum to a modern age I think.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8I always see these questions like "find the number of roots of this polynomial" or "show this ridiculous trig identity is true without any geometric picture that it came from" ...why?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8It's like teachers want people to believe math is useless AND ugly.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Whut, qwerty? The topranker got 360?! Surely the questions must have been really simple. The topper at my coaching (me) got 307/360 which was an exception to the rule that the previous years' toppers got no more than 250260.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8lol duuuude gotta go for that \(2 \pi\) score man!

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2the paper wasn't enough easy to get 360/360. the guys are too crazy...they are like 24 x 7 studying.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Their parents must be awful people

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Ok ok maybe I'm being pessimistic here haha

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2The guy who got 1st rank is a vry good frnd of mine. He says his parents think that he can do whatever he wants but he must remain healthy eating proper food.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2his parents don't interfere his personal life

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Woah he must be a genius :O

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2this guy got 1st rank in 2/4 test held till date

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Kaunsa institute hai ye?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1see you cant study more than 14 hours each day, you need at least 6 hours of sleep and 4 hours for maintenance

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Tumhe internet mil jaata hai?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2:) badiya bhi hai aur thodi prblems bhi hai....roz ghar par koi na koi guest aa jate hai...roz tv etc etc ka shore distrb karta hai...nd many more prblems

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Guest kyun aa jaate hain?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2pata nai... btw mere relatives paas me hi rahte hai to wo dadi dada se milne aa jate hai.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2btw @ganeshie8 did u got the answer to the question  dIzI/dz z=complex number

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1common sense : common lannguage guys, when others are there in the thread

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1thats okay, just saying..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3あなたは何について話していますか？理解できません。

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dIzI/dz derivative doesnt exist

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3What kind of derivative is this?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1z is a complex number f(z) = z is a real valued function that takes complex number as input

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, I can see that. But it's a complex number. How do you compute derivatives of complexvalued functions? Is it that instead of taking a small interval, you take a small circular area?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Exactly, the "interval on real number line" becomes a "ball in argand plane"

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3@imqwerty knows all this?! o.O

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Well, there goes my selfesteem.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1if you're in 12th grade and if you say you know all this, then you're just lying

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3He's in 11th grade!

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2i asked my teacher to explain this but he refused :( he said  its of no use

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3That's stupid. "It's not there in the JEE examination" should not be restated as "it's of no use".

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I think complex derivatives are a hell lot more useful that the four questions you gave us.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2yes!! the teaching methodology nd the mindset of ppl must change.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1hmm your teacher knows more than you, he might be thinking that teaching you about complex derivatives is useless because you ppl are not ready yet

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3What if he straightup doesn't know what they are? That's a possibility too. I don't know much about Allen teachers but I don't expect mine to give a satisfactory explanation.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2es this can be the reason @ganeshie8 nd we nt taught complex numbers till nw ..but still he said that its far away frm JEE syllabus. but i'll ask him when the chapter gets over. I asked Anna sir [HOD math allen] nd m sure he knws it

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Haha, of course it's beyond the reaches of JEE syllabi. It's not just the complex numbers we're taught in 11th and 12th.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Wow, I just searched him up and he looks pretty young, like a passout.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2m enrolled in vibrant academy too i'll go and ask Vikas gupta sir (HOD math vibrant).i think he will gimme the explanation.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I think it's better to not ask for it...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1If you get a satisficatory explanation, please do share wid us

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2haahaha once i got kicked out (frnkly) fr asking such doubts.. nd yes i'll definately share it.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2pk have u joined any institute?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider \(x^2+x+1\) divides into both \(x^5+x+1,x^{25}+x^5+1\). this is sufficient to prove \(5^{5^{n+1}}+5^{5^n}+1\) is not prime for positive integer \(n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the next problem, you just break it into an iterated sum, though if you don't set a lower bound on \(i,j\) individually you will run into problems

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2thanks @oldrin.bataku heres the new question that m solvin 

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Eh, this problem isn't that hard.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3It may be calculative, but it's not hard.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1looks that fraction evaluates to \(\left(\dfrac{n+1}{n}\right)^2\)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2Yes it has lil bit calculation prt

Empty
 one year ago
Best ResponseYou've already chosen the best response.8You can do the gauss fliperoo method there I think.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\sum_{k=1}^n (2k)^2=4\sum_{k=1}^n k^2=4\cdot\frac{n(n+1)(2n+1)}6\\\sum_{k=0}^n(2k+1)^2=\sum_{k=0}^n (4k^2+4k+1)=4\sum_{k=1}^n k^2+4\sum_{k=1}^n k+\sum_{k=0}^n 1$$ so the second sum gives us: $$4\cdot\frac{n(n+1)(2n+1)}6+4\cdot\frac{n(n+1)}2+(n+1)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, the second should be the sum up to \(n1\), actually

Empty
 one year ago
Best ResponseYou've already chosen the best response.8@imqwerty You might like Mathematical Methods in the Physical Sciences by Mary Boas. I think you might like it, it introduces complex numbers and complex analysis in separate chapters, check it out: https://faculty.psau.edu.sa/filedownload/doc4pdf0a187866618ca3049030ec5014860ae8original.pdf It's very accessible and if you have any questions feel free to ask, that book covers so much more though than just that.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2ok i am getting 151 as the answer

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2thank u so much @Empty ^^

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2wait no m getting 150 as the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so $$\frac23\cdot n(n1)(2n1)+2 n(n1)+n$$ which reduces the rational expression as $$\frac{2n+2}{2n1}>1.01\\2n+2>1.01(2n1)\\2+1.01>2(1.011)n\\3.01>0.02n\\n<50\cdot 3.01=150.5$$ so \(n\le 150\)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2correct @oldrin.bataku :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the last one \(xyz+xy+yz+zx\) links us to $$(x+1)(y+1)(z+1)=(x+1)(yz+y+z+1)\\\quad =xyz+xy+zx+yz+x+y+z+1$$ so we ahve that $$(x+1)(y+1)(z+1)(x+y+z)1=xyz+xy+zx+yz$$ now, we know \(x+y+z=12\) so: $$(x+1)(y+1)(z+1)13=xyz+xy+zx+yz$$ now, let's use the AMGM on \(x+1,y+1,z+1\): $$\frac{(x+1)+(y+1)+(z+1)}3\ge \sqrt[3]{(x+1)(y+1)(z+1)}$$ so it follows: $$(x+1)(y+1)(z+1)\le \left[1+\frac{x+y+z}3\right]^3$$ which gives us: $$(x+1)(y+1)(z+1)\le \left[1+4\right]^3=125$$ so we have that $$(x+1)(y+1)(z+1)13\le12513=112$$

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2thank u so much!! @oldrin.bataku now all the questions r solved ^~^

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2@Empty in the second question in the last step u took the common ratios of the 2 infinite GPs wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0complex differentiation is not beyond what a high schooler can know; I used the residue theorem in calculus II which was in 11th grade

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyways, \(z=z\bar z\) is not differentiable on \(\mathbb{C}\) because it depends on \(\bar z\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\sum_{k=0}^N\left(\frac{3^{k+1}}{2^{k+1}  3^{k+1}}  \frac{3^k}{2^k  3^k}\right)=\frac{3^{N+1}}{2^{N+1}3^{N+1}}\frac{3^0}{2^03^0}$$ the first term behaves like: $$\frac{1}{(2/3)^{N+1}1}\to \frac1{1}=1$$as \(N\to\infty\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also it should be \(k=1\) giving: $$\frac{3^1}{2^13^1}=\frac{3}{23}=\frac{3}{1}=3$$ so \(1+3=2\)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2thanks @oldrin.bataku ^~^ :)
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