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PrithiviRaj
PrithiviRaj
02:41
@Brahadeesh Did they ask only 2-3 questions in IIT B or IIT K interview ??
user185131
03:03
@PrithiviRaj No, around 5-6 questions actually
user185131
03:16
@ManeeshNarayanan all the very best! :)
PrithiviRaj
PrithiviRaj
@Brahadeesh okay ! thanks
Maneesh Narayanan
Maneesh Narayanan
03:40
@Brahadeesh Thank you very much.
BAYMAX
BAYMAX
03:50
Hi chat!!
Captain Bohemian
Captain Bohemian
04:13
I am in hunger trouble now.
user image
 
9 hours later…
vidyarthi
vidyarthi
12:58
@Brahadeesh hello. did you not write CMI Exam? What about IISc interview?
user185131
13:20
@vidyarthi hello
user185131
I did not write the CMI exam
user185131
IISc interview I gave, but it was pathetic :P
user185131
I didn't write the CMI exam because it was clashing with IIT Kanpur's interview, and I had given my centre as Mumbai, and the people at CMI refused to change my centre.
vidyarthi
vidyarthi
13:35
@Brahadeesh what questions were asked at IISc?
 
2 hours later…
user185131
15:07
They asked how many field automorphisms of R are there, and then asked me to prove my answer. There were some toological questions which I don’t remember. Some Galois theory question based on finite fields that I don’t remember. If A^2 = A then is A diagonalisable?
Maneesh Narayanan
Maneesh Narayanan
If A^2 = A then is A diagonalisable? I can do this only :(. did you say your interesting subject is topology and algebra?
0
Q: how to show that $\sum_{ n=1}^{∞ }a_n^+ = ∞ $?

user469754Let $a_n \in \mathbb R$, such that $\sum_{ n=1}^{\infty }|a_n| = \infty $ and $\sum_{ n=1}^{m }a_n \ \rightarrow a\in \mathbb R$ as $m \rightarrow \infty$. Let $a_n ^+= \max\{a_n,0\}.$ Show that $\sum_{ n=1}^{\infty }a_n^+ = \infty $. My attempts : I was thinking like this $$|a_n|=a_n^{+}...

real-analysis
@Brahadeesh did you solve this question for IITB exam preparation? can you give hint? OR you can type answer there :)
It seems like $\sum(-1)^n \frac{1}{n}$
user185131
15:38
@ManeeshNarayanan Yeah, I said my interests are algebra and topology
Maneesh Narayanan
Maneesh Narayanan
okay.
user185131
@ManeeshNarayanan you're right, it is of the same form. Suppose the positive terms sum to a finite number. Then, what happens when you write down \sum |a_n|? Since this is a sum of only positive terms, we can reorder it in any way we like and the value of the sum will not change. So, let's write it as \sum |a_n| = \sum |<positive a_n's>| + \sum |<negative a_n's>|
user185131
The LHS is infinity (given). The first term on the RHS is finite by assumption. So, \sum |<negative a_n's>| is also \infty. This means that \sum <negative a_n's> is -\infty, or in other words, the partial sums of the negative terms is unbounded. Now, this implies that \sum a_n is unbounded, but it was given to be bounded, which is a contradiction.
Maneesh Narayanan
Maneesh Narayanan
Thank you @Brahadeesh
user185131
16:21
No problem! :)

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