Recent content by kendricklamarlover101

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u guys are tweakin every sdehs task ive seen is harder than any putnam exam and the imo. not even terrence tao could score above a 50% these state rankers are a different breed.

shouldnt it be k^8 det(A) as when a row/column is multiplied by a scalar k then det(A) increases by a factor of k? so since its an 8x8 matrix it would be k^8?

u can write det(WM^-1) as \det (-4MM^{T}M^{-1}) = -4\det(MM^{T}M^{-1}) =-4\det(M)\det(M^{T}M^{-1}) =-4\det(M)\det(MM^{-1}) = -4\det(M) u could also split the product up at the start to get \det (-4MM^{T}M^{-1}) = -4\det(M)\det(M^{T})\det(M^{-1}) =-4\det(M)\det(M)\frac{1}{\det(M)} =-4\det(M)

u can consider cases as |x+1| = -(x+1) for x< -1 and |x-5| = -(x-5) for x < 5. then from there u just solve the inequality for each case (x<-1, -1<x<5 , x > 5) for example -1<x<5 we have: x+1-(x-5) > 7 => -4 > 7 which isnt true so theres no value of x between -1 and 5 that satisfies the...

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for c) the harmonic series diverges so it is bounded above and the lower bound will just be 1

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one way to approach it is using the sum to product identity (sin(A+B) - sin(A-B) = 2cosAsinB).

the both have a common factor of (n-r-2)! so u can just multiply it by (n-r-1) like how 1/3 + 1/6 = 2/6 + 1/6 instead of 6/18 + 3/18

im pretty sure u can use the heaviside cover up method for partial fractions without proof

lol these emojis are so silly like has anybody ever used :chainsaw2:

yup 👍

yup basically just adding that limit notation to show that ur actually taking the infinite sum

if u do take the limit u get that the result would equal 1 + ln(2) + ln(2) > 1+ln2 u can take the limit from the start just by writing \lim_{k \rightarrow \infty} \sum_{n=2}^{k} \ln \left( \frac{n^2}{n^2-1} \right ) and if u want to expand the sum out to see the telescoping series the result...