1,516 reputation
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bio website cse.iitb.ac.in/~aruniyer
location India
age 30
visits member for 2 years, 2 months
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May
16
comment How to prove such an elementary inequality
Precisely. (Surely you meant pth root instead of pth power).
May
14
comment $6^{(n+2)} + 7^{(2n+1)}$ is divisible by $43$ for $n \ge 1$
Yes, that is the induction hypothesis step. But choose some other alphabet other than n, since it is already being used to represent the power. Say P(k) = 43*m, where m >= 13. Substitute that and you will get the common factor of 43.
May
14
comment $6^{(n+2)} + 7^{(2n+1)}$ is divisible by $43$ for $n \ge 1$
Note that LHS is P(k+1) and RHS is $6\times P(k) + 43\times7^{2k+1}$. What do you know about P(k)?
Apr
8
comment orthogonal complement problem: show $\operatorname{oc}(A\cap B)=\operatorname{oc}(A)+\operatorname{oc}(B)$
Oh I missed the part of $A$ and $B$ being subspaces. Thanks.
Mar
31
comment Using SVD in PCA for image compression
You seem to be confusing scaling with compression. If you just want to reduce the dimensions, then just downscale your images. en.wikipedia.org/wiki/Image_scaling
Mar
27
comment Set of vectors linearly independent
Note that $u_i = v_i - v_{i - 1}$ for $i > 1$
Mar
27
comment finding the unspecified ${\bf E}[X]$ and $\rm var(X)$ given the expectation of higher powers of $X$
A standard normal random variable has mean 0 and variance 1.
Dec
31
comment Relation between convex set and convex function
$\phi(\lambda x + (1 - \lambda)y)=\inf_{a\in A}\|\lambda x + (1 - \lambda)y-a\|$. Think of how you can play around with RHS. Note that, $\lambda a + (1 - \lambda) a = a$.
Dec
3
comment Proving by induction that $1+\frac{1}{2}+\frac{1}{3}+…+\frac{1}{n}\le\frac{n}{2}+1$ holds for all $n \ge 1$
?? Haven't they simply added 1/n+1 to both sides of the hypothesis statement. Why is it so magical to you?
Nov
24
comment $ \begin{bmatrix} -2 & 5 & 4 \\-1 & 0 & 0 \\0 & 4 & 3 \end{bmatrix}^{2013}$ =?
Sure, give me sometime and I will actually do it and write it up. Do you want me to? My point that it was doable precisely because it is a special matrix.
Sep
10
comment Support of a vector
support of a vector is the number of non-zero elements in that vector.
Sep
6
comment What is the difference between commutatitivity and distributivity?
No, nothing to add. Hurkyl (+1), pretty much has said everything I wanted to convey.
Sep
6
comment What is the difference between commutatitivity and distributivity?
Do you understand my question? Given two operators A and B, AB = A(B(x)). Unless B's range is A's domain, A(B(x)) does not make sense. Similarly, BA = B(A(x)) therefore, A's range has to be B's domain. Finally, for AB = BA to be true, A's range has to be same as B's range. Thereby, unless you define A and B's domain and range, you cannot talk about commutativity. That is why, I asked you, please write the domain and range of the two operators that you speak. Then we can think about commutativity.
Sep
6
comment What is the difference between commutatitivity and distributivity?
Please define the domain and range of your operators A and $\sum$. Without that, talking about commutativity is pointless.
Aug
27
comment Counting Question
Choose the even number first, then select the other numbers.
Aug
27
comment Reproducing Kernel Hilbert Spaces for Dummies
That's pretty much what it seems to be if I read that paragraph in the paper correctly.
Aug
25
comment How to show that $\{\sqrt m - \sqrt n : m,n \in \mathbb N\}$ is dense in $\mathbb R$?
math.stackexchange.com/questions/275047/…
Aug
25
comment solving equations by the method of elimination
@rahul, now multiply throughout by $6x(x+1)$, what do you get?
Aug
25
comment solving equations by the method of elimination
@rahul, are you sure your final equation is correct? Please check your equations again.
Aug
24
comment Metric spaces and openness
a] math.stackexchange.com/questions/203728/… , b] Given any set X, consider all open sets in the metric space that contain X (call them $G_\alpha, \alpha \in A$ where A is some index set) and let $Y = \cap_{\alpha \in A} G_\alpha$. Show that $Y = X$.