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For each of these functions, find the least integer n such that f(x) is Ox). If E = (24+22+1) (z+1) then the least integer n is

For each of these functions, find the least integer n such that f(x) is Ox). If E = (24+22+1) (z+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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This is a multi-part question. e an answer is su 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). of 2 NO k ces If 8 (x²+z²+1) (24+1) then the least integer nis
Transcribed Image Text:This is a multi-part question. e an answer is su 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). of 2 NO k ces If 8 (x²+z²+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 nnk = nk+1. We observe that all the integers from 1 to n are at most n, hence 1+2++ n ≤ n + n + ... + nk. Step 3 We can simplify the sum by adding the exponents: nk+n++ nk = nk+k+-+k = nnk Step 4 We observe that all the integers from 1 to n are at most n, hence 1k+2k++n* ≥nk +nk + ... + n*. By combining our calculations, we get 1* + 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). By combining our calculations, we get 1k + 2k + ... + n ≤ n k+1 for all n ≥1. We have verified the definition of what it means for 1k+2k++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: n* + n* +...+n* = 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 nnk = nk+1. We observe that all the integers from 1 to n are at most n, hence 1+2++ n ≤ n + n + ... + nk. Step 3 We can simplify the sum by adding the exponents: nk+n++ nk = nk+k+-+k = nnk Step 4 We observe that all the integers from 1 to n are at most n, hence 1k+2k++n* ≥nk +nk + ... + n*. By combining our calculations, we get 1* + 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). By combining our calculations, we get 1k + 2k + ... + n ≤ n k+1 for all n ≥1. We have verified the definition of what it means for 1k+2k++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: n* + n* +...+n* = n⋅nk.
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