Why is identity theft exploding in developing countries? – Everything else

As indicated by 2018, Identity Guard reports fraud: Fraud leads to a new era of complexity of Javelin Strategy and Research. There were 16.7 million identity theft victims in 2017, a record high that reached a record high last year. Criminals engage in complex conspiracy to blackmail identities, leaving record numbers of exploited. The stolen sum reached $ 16.8 billion a year ago, than 30 percent of the US

Be careful on the internet:

In the UK, the majority of the population uses broadband services. The internet makes everything easy for the buyer. Half of the UK population uses online account management, pays their bills, buys and uses the Internet to perform their daily tasks.

Precautions against identity theft:

  • As per Identity Guard reviewsNever share your social security number unless it is absolutely necessary. Also, do not bring your social security card with you anywhere.
  • It is said, always try to separate personal information as much as possible, because if something has lost, at least strangers have only one piece of information and not everything.
  • If you want to communicate with your bank through social networking sites such as the live chat support option, never include your bank details in your text message.
  • Never enter a real age and places on Twitter, Facebook and other social media apps.
  • Before you publish information such as photos, videos and other details, you should search for the link that says privacy and policies.
  • Never use not licensed and expired spyware, firewalls, and other antivirus programs because they are not eligible for software updates.
  • Always create a strong and unique password for your social media account. How do you choose a password that contains both letters and numbers.
  • Do not sign in to your social media account from public Wi-Fi and computers.
  • Watch what links you click, like and share.
  • If you have experienced, you are a victim of identity threats, without much thought, share your difficulties with the cybercrime Agencies. More information can be found here: www.noidentitytheft.com

What is a general rule for this small lambda calculus identity?

I messed around with a project that does bring about some normalization of lambda-calculus-like expressions, and I found it

(λ λ ... λn (n-1) ... 2 1)

(De Brujin notation) is just the identity. I have generalized this a bit to realize that I can do such things

(λ λ λ λ 4 2 1) = (λ λ 2)

Basically, if you can identify a character in your expression that is just the variable index nand all variable indexes to the left of this point are larger than nand to the right, it's all down to the variable indices to index 1 Then you can remove all indices to the right of this character (including this character) and then the _λ_s from the head of the expression and reduce the remaining indices accordingly.

This feels too arbitrary to be the simplest form of this identity. There has to be a cleaner way to express this, and possibly a more general rule.

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Binomial Coefficients – Pascal Triangle and Hockey Stick identity verification using combinatorics

How would I prove the following with combinatorial proof?

(a) Show that this identity is in Pascal's triangle:
$$ sum_ {k = 0} ^ {n} binom {n} {k} ^ {2} = binom {2n} {n}, ∀n ∈ mathbb {N} $$

(b) Prove the hockey stick identity:
$$ sum_ {k = 0} ^ {m} binom {n + k} {k} = binom {n + m + 1} {m}, m, n mathbb {N} text {with} m ≤ n $$

Number Theory – A Surprising Identity: $ det[cospifrac{jk}n]_ {1 le j, k le n} = (- 1) ^ { lfloor frac {n + 1} 2 rfloor} (n / 2) ^ {(n-1) / 2} $

Based on my calculation, I present my following conjecture regarding the cosine function.

guess, For every positive integer $ n $We have the identity
$$ frac1 {2n} det left[cospifrac{jk}nright]_ {0 le j, k le n} = det left[cospifrac{jk}nright]_ {1 le j, k le n} = (- 1) ^ { lfloor frac {n + 1} 2 rfloor} (n / 2) ^ {(n-1) / 2}. $$

This is part of Conjecture 5.7 in my preprint arXiv: 1901.04837. The paper contains similar assumptions.

Ideas for a solution of guesswork?

Linear algebra – Can a matrix not be a multiple of identity, have repeated eigenvalues ​​and still be diagonalizable?

The question: The diagonalizability of 2 × 2 matrices with repeated eigenvalues ​​suggests that a matrix whose eigenvalues ​​are different must be diagonalizable. At a multiple of the identity matrix, however, all eigenvalues ​​are equal and yet diagonalizable. I suspect it's a general $ n times n $ Matrix is ​​not a multiple of the identity (has some non-zero diagonal elements) and has a variety of eigenvalues, it is not diagonalizable. I could not find a counter example to it. Can it be proved (or refuted with a counterexample).