Hello can anyone do the work for this question (IT SAYS QUESTION 10) needed that

Question

Hello can anyone do the work for this question (IT SAYS QUESTION 10) needed that's based off of the style of these notes that I posted under it? I'd greatly appreciate it thanks.. If the photo shows small please open the images in a separate tab for it to show bigger please. The first image is the question i need answered and the photos under it are the notes style. Please realize that the notes are not the same exact questions.. but gives an idea. Let me know if I need to update the question somehow please and thanks.

D Question 10 Use formal definitions to show that: n4 + n-500 = (n2) 4n2-n-1000- (n*) Show your work, similar to the examples from the notes.

Explanation / Answer

1.

For n4 +n -500 = (n2) ,there exists a positive real constant c and there exists an integer constant n0 >= 1 such that n4 +n -500 >= c*(n2)  for every integer n >= n0.

Now , n4 +n -500 >= n4 for n >= 500

and , n4 >= n2 for n >= 500

So n4 +n -500 >= n4 >= c*n2 for c=1 and n >= 500

Thus n4 +n -500 = (n2) for n0 = 500 and c=1.

2.

For 4*n2 -n -1000 = (n2) ,there exists two positive real constant c1 and c2 and there exists an integer constant n0 >= 1 such that  

c1 * n2 <= 4*n2 -n -1000 <= c2 * n2

Now , c1* n2 <= 4*n2 -n -1000 = 2 * n2 + (n2 - n) + (n2 - 1000) for c1 = 1 and n >=1000

because , (n2 - n) > 0 , (n2 - 1000) > 0 and 2 * n2 > c1* n2

for n >=1000 and c1 = 1

Also ,  4*n2 -n -1000 <= c2 * n2 for c2 = 4 and n >= 1

So, there exists two positive real constant c1 = 1 and c2 = 4 and a positive integer constant n0 >= 1000

such that  

c1 * n2 <= 4*n2 -n -1000 <= c2 * n2 for c1 = 1,c2 = 4 and n >= n0 = 1000

3.

For n3 - 3*n + 15 = O (n4) ,there exists a positive real constant c and a positive constant n0 such that n3 - 3*n + 15 <= c*(n4)  for n >= n0.

Now ,  n3 - 3*n + 15 <= n3 for n >=5

and, n3 <= c*n4 for c=1 and n >=1

So, n3 - 3*n + 15 <= n3 <=  c*n4 for c=1 and n>=5

Thus,  n3 - 3*n + 15 <=  c*n4 for c=1 and n >= n0 = 5

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