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Apr
26
answered How to deduce the formula “distribution” in groups? What is the difference between “distribution” & “arrangement”?
Mar
16
comment Stirling number of the second kind and combinations
Actually i wanted to create a function which maps element from the (n-r)-sized subset to the set of r-partitions. If you have a function like this which is either onto or one-one then you are done with your proof.
Mar
16
answered Stirling number of the second kind and combinations
Feb
6
comment Loss functions for regression
You have used x and x' both , so i was little confused.
Feb
6
comment Loss functions for regression
Thanks for your answer. However, i have a doubt. I would be grateful if you could clarify that. Here, y(x) is the predicted value for x . And in the training data expected input-output pairs <x,t> are give. Both x and y(x) are real-valued. So, i am not sure why we need x' in the computation?
Feb
5
asked Loss functions for regression
Feb
1
comment Induction Proof Check: For a binary tree T, Prove that the number of full nodes in T is always one less than the number of leaves in T.
you can't use the recursion for F directly. When you have the recursion you can simply unwind it in this way F(n) = n-2 + F(2) = n-1 to obtain the answer, without using induction. On the other hand by proving the fact by induction you obtain F(n) = n-1. Also note F(n) is defined for n=1.
Feb
1
answered Induction Proof Check: For a binary tree T, Prove that the number of full nodes in T is always one less than the number of leaves in T.
Jan
26
answered Histogram of duplication in n choose k
Jan
8
revised Representing “not” in lambda calculus
added 357 characters in body
Jan
7
comment Representing “not” in lambda calculus
what i tried to mean is f (or P, i just edited) does not have any more lambda inside it.
Jan
7
revised Representing “not” in lambda calculus
edited body
Jan
7
asked Representing “not” in lambda calculus
Jan
6
comment Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s
yes, it is. Thanks for pointing.
Jan
6
revised Use the PIE to prove an identity
added 2 characters in body
Jan
6
revised Use the PIE to prove an identity
added 133 characters in body
Jan
6
asked Use the PIE to prove an identity
Jan
6
revised Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s
added 240 characters in body
Jan
6
revised Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s
Slight correction in the formula
Jan
6
revised Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s
added 12 characters in body