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  • Mathematics Stack Exchange
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  • How to prove the divisibility rule for $3\\, $ [casting out threes]
    The divisibility rule for 3 is well-known: if you add up the digits of and the sum is divisible by 3, then n is divisible by three This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule: 1212582439 → 37 → 10 → 1 3 ∤ 1212582439
  • Methods for choosing $u$ and $dv$ when integrating by parts?
    When doing integration by parts, how do you know which part should be u u ? For example, For the following:
  • Taylor series of $\\ln(1+x)$? - Mathematics Stack Exchange
    You got the general expansion about x = a Here we are intended to take a = 0 That is, we are finding the Maclaurin series of ln(1 + x) That will simplify your expression considerably Note also that (n − 1)! n! = 1 n The approach in the suggested solution also works We note that 1 1 + t = 1 − t + t2 − t3 + ⋯
  • epistemology - Fallacy by Sherlock Holmes Eliminate the impossible . . .
    Well, the fallacy would not be in Sherlock Holmes line; that remains perfectly valid The fallacy would be in the hybris of the person who did not carefully conduct an exhaustive search for alternatives In order to use "whatever remains, however improbable, must be the truth" you must exhaust the space of possibilities first If you didn't do that, you are not entitled to appeal to Sherlock
  • linear algebra - Does $\det (A + B) = \det (A) + \det (B)$ hold . . .
    Can there be said anything about det (A + B)? If A B are symmetric (or maybe even of the form λI) - can then things be said?
  • Justifying why 0 0 is indeterminate and 1 0 is undefined
    0 0 = x 0 0 = x 0x = 0 0 x = 0 x x can be any value, therefore 0 0 0 0 can be any value, and is indeterminate 1 0 = x 1 0 = x 0x = 1 0 x = 1 There is no such x x that satisfies the above, therefore 1 0 1 0 is undefined Is this a reasonable or naive thought process? It seems too simple to be true
  • limit when zero divided by infinity - Mathematics Stack Exchange
    I know that limx→∞ f(x) = 0 lim x → ∞ f (x) = 0 and limx→∞ h(x) = ∞ lim x → ∞ h (x) = ∞ So at the and I have 0∞ 0 ∞ I know that infinity is not a real number but I am not sure if the limit is indeterminate (Also, there are people who are saying contradictory things on internet) I know very well that it is not possible to use Hopital's rule My guess is that : As we





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