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which q
Last one x = y = z = 4

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any question that u like...these r the question which i wsn't able to solve
First 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
@ParthKohli how did u get x=y=z=4
It's a symmetric situation. Apply all inequalities you please, but you'll finally get to this.
Simplest way to think of it (not a proof) is that \[n^2 > (n-a)(n+a)\] when a is not equal to zero, so that's the max when they're all the same.
first q doesnt look easy
I think we can split that 3rd one up using partial fractions, then it becomes a telescoping series that way.
Good questions by the way!
thanks @Empty @ParthKohli i can't get to the result x=y=z. can u tell how to start ..
I'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.
Ok so for the second one I'm getting: $$\frac{1}{1-9}\frac{1}{1-4}\frac{5}{6}$$ Does that match your answer?
wait m doing the 1st one
for 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
Yeah 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: $$(a^2+a+1)(a^3-a^2+1)$$
@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.
@Empty we can't write 5^5^(n+1) as a^5 if a=5^5^n
OK, 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.
yes so how do we know that in which case x=y gives the max
$$5^{5^{n+1}}=5^{5*5^n} = (5^{5^n})^5$$ @imqwerty
You 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.
i feel #1 is still a stupid q, you're not gona learn anything frm it
wait sry yes we can put a^5...
yes it is @ganeshie8 :D
Y'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.
Ok 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!
Isn't that the standard way to do it though? How did you originally do it?
I factored it
the 1st one is solved thanks guys :)
\[\begin{align} x^5 + x + 1 & = (x^5-x^2)+(x^2+x+1)\\ \\ & = x^2(x^3 - 1) + (x^2 + x + 1) \\ \\ & = x^2(x-1)\color{blue}{\bf (x^2 + x + 1)} + \color{blue}{\bf (x^2 + x + 1)}\end{align}\] http://math.stackexchange.com/questions/477295/factor-x5-x-1
To be fair, that solution was most likely obtained by working backwards. :P
that is okay for a proof
Yes of course, but you can't really think of that on your own.
that all goes as "scratch work" :P
ahhh 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
I'm ashamed that I have no other ideas other than telescoping for #3.
It wants to telescope, why force it to be anything it's probably not? :P
When you separate it out, the coefficients are both 1, it's truly a beautiful result
I'm either dumb or too much drunk to notice it but why do you think telescoping is a bad idea ?
Yeah you're not drunk enough ganeshie, cause this baby wants to see the moon
I 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.
m really confused nd annoyed nd kinda drunk ..cheers @ganeshie8 @Empty ..can u please tell me what u did with the 2nd question
my mind is jst nt wrkin normal today :(
OK, it's actually telescoping really nicely.
Here 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) \[\left(\sum_{i=0}^{3} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{3} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{3} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{3} \frac{1}{2^{2j+1}} \right) \]
no i mean hw did u split that
No, I _created_ this
so 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
\[\frac{6^k}{(3^k - 2^k)(3^{k+1 } - 2^{k+1})}\]\[= \frac{3^{k+1}}{2^{k+1} - 3^{k+1}} - \frac{3^k}{3^k - 2^k}\]
There's a typo.
|dw:1438453754914:dw| So this is the first half of the terms! @imqwerty
Supposed to be\[= \frac{3^{k+1}}{2^{k+1} - 3^{k+1}} - \frac{3^k}{2^k - 3^k}\]
Specifically 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.
so that's where the 2i and 2i+1 is coming from, keeping the powers even and odd: \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) \]
thank u so much @Empty :)
OK, so we're sorta done with all of the questions.
whats the ansur for q2
Are you asking me or asking imqwerty?
m calculating the answer...
@Empty Did you check out the solution to the problem I gave you yesterday?
Oh no I didn't, also I ended up getting 2 to that, I'll go check real fast
2 is correct.
:) a big THANKS TO ALL OF YOU @Empty @ParthKohli @ganeshie8 :)
#2 im getting 5/144
Ganeshie 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. \(F_1 = F_2 = 1 \).
my answer the #2 is 121/144
I am familiar with that problem \[\sum_{n=1}^{\infty}F_nx^n= \frac{x}{1-x-x^2}\] plugin \(x=1/2\)
Here's the solution: \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) \] \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \frac{1}{3}\left(\sum_{j=0}^{\infty} \frac{1}{3^{2j}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \frac{1}{2} \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j}} \right) \] \[\left( \frac{1}{3} + \frac{1}{2} \right) \left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j}} \right) \] \[\left( \frac{1}{3} + \frac{1}{2} \right) \left( \frac{1}{1-3^2} \right) \left(\frac{1}{1-2^2} \right) \] \[\frac{5}{144}\]
damn 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 :(
Haha 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
I'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
sometimes 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 ?
they just want the exams to be tough, they dont bother to look at the syllabus of 12th grader
the 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 :(
at my home m nicknamed - 'silly mistake'
trust me, you and PK are way better than me when i was in my 12th
Yeah I don't understand the obsession with rote memorizing all these useless one-time-use tricks for problems mankind invented calculators and wolfram alpha for. Time to change the curriculum to a modern age I think.
@Empty so true^ .
I 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?
It's like teachers want people to believe math is useless AND ugly.
Whut, qwerty? The top-ranker 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 250-260.
humblebrag
lol duuuude gotta go for that \(2 \pi\) score man!
the paper wasn't enough easy to get 360/360. the guys are too crazy...they are like 24 x 7 studying.
Their parents must be awful people
Ok ok maybe I'm being pessimistic here haha
The 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.
his parents don't interfere his personal life
Woah he must be a genius :O
this guy got 1st rank in 2/4 test held till date
Kaunsa institute hai ye?
see you cant study more than 14 hours each day, you need at least 6 hours of sleep and 4 hours for maintenance
allen
Kota wala?
han
Tumhe internet mil jaata hai?
mai to kota ka hi hu
Oh, badiya yaar.
:) 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
Guest kyun aa jaate hain?
pata nai... btw mere relatives paas me hi rahte hai to wo dadi dada se milne aa jate hai.
btw @ganeshie8 did u got the answer to the question - dIzI/dz z=complex number
common sense : common lannguage guys, when others are there in the thread
ok :) sry
thats okay, just saying..
あなたは何について話していますか?理解できません。
dIzI/dz derivative doesnt exist
What kind of derivative is this?
z is a complex number f(z) = |z| is a real valued function that takes complex number as input
Yeah, I can see that. But it's a complex number. How do you compute derivatives of complex-valued functions? Is it that instead of taking a small interval, you take a small circular area?
Exactly, the "interval on real number line" becomes a "ball in argand plane"
@imqwerty knows all this?! o.O
not too much..
Well, there goes my self-esteem.
if you're in 12th grade and if you say you know all this, then you're just lying
He's in 11th grade!
So am I.
Me too...lol
i asked my teacher to explain this but he refused :( he said - its of no use
That's stupid. "It's not there in the JEE examination" should not be restated as "it's of no use".
^true that
I think complex derivatives are a hell lot more useful that the four questions you gave us.
yes!! the teaching methodology nd the mindset of ppl must change.
hmm 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
What if he straight-up 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.
es 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
Haha, of course it's beyond the reaches of JEE syllabi. It's not just the complex numbers we're taught in 11th and 12th.
Wow, I just searched him up and he looks pretty young, like a passout.
m enrolled in vibrant academy too i'll go and ask Vikas gupta sir (HOD math vibrant).i think he will gimme the explanation.
I think it's better to not ask for it...
If you get a satisficatory explanation, please do share wid us
haahaha once i got kicked out (frnkly) fr asking such doubts.. nd yes i'll definately share it.
pk have u joined any institute?
Yeah, Vidyamandir.
oh nice :)
consider \(x^2+x+1\) 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 \(n\)
for 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
thanks @oldrin.bataku heres the new question that m solvin -
1 Attachment
Eh, this problem isn't that hard.
It may be calculative, but it's not hard.
looks that fraction evaluates to \(\left(\dfrac{n+1}{n}\right)^2\)
Yes it has lil bit calculation prt
You can do the gauss fliperoo method there I think.
$$\sum_{k=1}^n (2k)^2=4\sum_{k=1}^n k^2=4\cdot\frac{n(n+1)(2n+1)}6\\\sum_{k=0}^n(2k+1)^2=\sum_{k=0}^n (4k^2+4k+1)=4\sum_{k=1}^n k^2+4\sum_{k=1}^n k+\sum_{k=0}^n 1$$ so the second sum gives us: $$4\cdot\frac{n(n+1)(2n+1)}6+4\cdot\frac{n(n+1)}2+(n+1)$$
oops, the second should be the sum up to \(n-1\), actually
@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/doc-4-pdf-0a187866618ca3049030ec5014860ae8-original.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.
ok i am getting 151 as the answer
thank u so much @Empty ^^
wait no m getting 150 as the answer
so $$\frac23\cdot n(n-1)(2n-1)+2 n(n-1)+n$$ which reduces the rational expression as $$\frac{2n+2}{2n-1}>1.01\\2n+2>1.01(2n-1)\\2+1.01>2(1.01-1)n\\3.01>0.02n\\n<50\cdot 3.01=150.5$$ so \(n\le 150\)
correct @oldrin.bataku :)
for 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 AM-GM 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 $$(x+1)(y+1)(z+1)-13\le125-13=112$$
thank u so much!! @oldrin.bataku now all the questions r solved ^~^
@Empty in the second question in the last step u took the common ratios of the 2 infinite GPs wrong
complex differentiation is not beyond what a high schooler can know; I used the residue theorem in calculus II which was in 11th grade
anyways, \(|z|=z\bar z\) is not differentiable on \(\mathbb{C}\) because it depends on \(\bar z\)
$$\sum_{k=0}^N\left(\frac{3^{k+1}}{2^{k+1} - 3^{k+1}} - \frac{3^k}{2^k - 3^k}\right)=\frac{3^{N+1}}{2^{N+1}-3^{N+1}}-\frac{3^0}{2^0-3^0}$$ the first term behaves like: $$\frac{1}{(2/3)^{N+1}-1}\to \frac1{-1}=-1$$as \(N\to\infty\)
also it should be \(k=1\) giving: $$-\frac{3^1}{2^1-3^1}=-\frac{3}{2-3}=-\frac{3}{-1}=3$$ so \(-1+3=2\)
thanks @oldrin.bataku ^~^ :)

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