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• Recall, ker(f) is a prime ideal if A₁ \ ker(f) is closed under multiplication. Suppose a,b ∈ A₁ are such that f(a),f(b) ≠ 0, i.e. a,b ∈ A₁ \ ker(f). Since A₂ has no zero divisors we have f(ab) = f(a)f(b) ≠ 0, meaning ab ∈ A₁ \ ker(f). → ker(f) …

  in abstract algebra rings? Comment by suleiman June 2021
• (n + 5)^5 = n^5 + 25n^4 + 250n^3 + 1250n^2 + 3125n + 3125 ≤ (1 + 25 + 250 + 1250 + 3125 + 3125)n^5 ∀n ⇒ (n + 5)^5 = O(n^5)

  in show (n+5)^5 is O(n^5)? Comment by suleiman June 2021
• (n + 5)^5 = n^5 + 25n^4 + 250n^3 + 1250n^2 + 3125n + 3125 ≤ (1 + 25 + 250 + 1250 + 3125 + 3125)n^5 ∀n ⇒ (n + 5)^5 = O(n^5)

  in show (n+5)^5 is O(n^5)? Comment by suleiman June 2021
• ∫_[0,1] {∫_[0,x] cos(x²) dy} dx = ∫_[0,1] {ycos(x²) |_[y=0,x]} dx = ∫_[0,1] xcos(x²) dx = ½sin(x²) |_[x=0,1] = ½sin(1)

  in How do you integrate cos(x^2)dydx? Comment by suleiman May 2021
• ∫_[0,1] {∫_[0,x] cos(x²) dy} dx = ∫_[0,1] {ycos(x²) |_[y=0,x]} dx = ∫_[0,1] xcos(x²) dx = ½sin(x²) |_[x=0,1] = ½sin(1)

  in How do you integrate cos(x^2)dydx? Comment by suleiman May 2021