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If fx 1(2) = (x+x²+1) (24+1) then the least integer n is

If fx 1(2) = (x+x²+1) (24+1) then the least integer n is

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please help me with these two questions. I am having trouble understanding what to do.

 

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f 2 k aces If This is a multi-part question. an answer is sub you will be unable to return to this For each of these functions, find the least integer n such that f(x) is O(x). 8 (x²+x²+1) (24+1) then the least integer nis
Transcribed Image Text:f 2 k aces If This is a multi-part question. an answer is sub you will be unable to return to this For each of these functions, find the least integer n such that f(x) is O(x). 8 (x²+x²+1) (24+1) then the least integer nis
Click and drag the steps that are required to prove 1k+ 2k +...+nk is O(nk + 1) to their corresponding step numbers. Step 1 Step 2 Using a law of exponentiation, we simplify n⋅nk=nk+1. Using a law of exponentiation, we simplify naknk+1 We observe that all the integers from 1 to n are at most n, hence 1k+2k++ n* ≤ n k +nk + ... + nk. Step 3 We can simplify the sum by adding the exponents: nk+n++ nk = nk+k+.+k = nrk. Step 4 We observe that all the integers from 1 to n are at most n, hence 1k+2k++nk ≥nk + n + ... + n*. By combining our calculations, we get 1 + 2k ++nk ≥nk+1 for all n ≥ 1. We have verified the definition of what it means for 1+2++nk to be O(n+1). By combining our calculations, we get 1k + 2k + + n ≤n*+1 for all n≥1. We have verified the definition of what it means for 1+2++nk to be O(n+1). Since there are exactly n equal terms in the sum, we can write the sum as a simple product: nk+n++nk = n⋅nk.
Transcribed Image Text:Click and drag the steps that are required to prove 1k+ 2k +...+nk is O(nk + 1) to their corresponding step numbers. Step 1 Step 2 Using a law of exponentiation, we simplify n⋅nk=nk+1. Using a law of exponentiation, we simplify naknk+1 We observe that all the integers from 1 to n are at most n, hence 1k+2k++ n* ≤ n k +nk + ... + nk. Step 3 We can simplify the sum by adding the exponents: nk+n++ nk = nk+k+.+k = nrk. Step 4 We observe that all the integers from 1 to n are at most n, hence 1k+2k++nk ≥nk + n + ... + n*. By combining our calculations, we get 1 + 2k ++nk ≥nk+1 for all n ≥ 1. We have verified the definition of what it means for 1+2++nk to be O(n+1). By combining our calculations, we get 1k + 2k + + n ≤n*+1 for all n≥1. We have verified the definition of what it means for 1+2++nk to be O(n+1). Since there are exactly n equal terms in the sum, we can write the sum as a simple product: nk+n++nk = n⋅nk.
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