Determine whether each of these proposed definitions is a valid recursive definition of a function ffrom the set of nonnegative integers to the set of integers. If fis well defined, find a formula for f(n) when n is a nonnegative integer and prove that your formula is valid. Prove by mathematical induction that the formula found in the previous problem is valid. First, outline the proof by clicking and dragging to complete each statement. Let P(n) be the proposition that f(n) = 1-n Basis Step: P(0) states that f(0) = 1-0 =1, which is the value given for f(0) in the definition of f Inductive Step: Assume that P(n) is true for all integers n ≥0 Show that f(k)=1-k for an arbitrary integer k≥ 0. We have completed the basis step and the inductive step. By mathematical induction, we know that Vk(P(k) → P(k+1)) is true, that is, Vk(f(k) 1-k→ f(k+1)=1-(k+1)) Vk(ƒ(k+1) = = IH = II Second, click and drag expressions to fill in the details of showing that VK(P(k)→ P(k+ 1)) is true, thereby completing the induction step. (You must provide an answer before moving to the next part.) (1-k) - 1 f(k) - 1 1- (k+1) P(n) is true for all integers n ≥0 f((k+1)-1)-1 Vk(P(k) → P(k + 1)) is true, that is, Vk(f(k)=1-k→ ƒ(k+1) = 1 − (k+1)) f(0) = 1-0 =1, which is the value given for f(0) in the definition of f f(k) = 1-k for an arbitrary integer k ≥ 0 f(n)=1-n 1- (k+1) f((k+1)-1)-1 (1-k) - 1 f(k) - 1 Reset Reset
Determine whether each of these proposed definitions is a valid recursive definition of a function ffrom the set of nonnegative integers to the set of integers. If fis well defined, find a formula for f(n) when n is a nonnegative integer and prove that your formula is valid. Prove by mathematical induction that the formula found in the previous problem is valid. First, outline the proof by clicking and dragging to complete each statement. Let P(n) be the proposition that f(n) = 1-n Basis Step: P(0) states that f(0) = 1-0 =1, which is the value given for f(0) in the definition of f Inductive Step: Assume that P(n) is true for all integers n ≥0 Show that f(k)=1-k for an arbitrary integer k≥ 0. We have completed the basis step and the inductive step. By mathematical induction, we know that Vk(P(k) → P(k+1)) is true, that is, Vk(f(k) 1-k→ f(k+1)=1-(k+1)) Vk(ƒ(k+1) = = IH = II Second, click and drag expressions to fill in the details of showing that VK(P(k)→ P(k+ 1)) is true, thereby completing the induction step. (You must provide an answer before moving to the next part.) (1-k) - 1 f(k) - 1 1- (k+1) P(n) is true for all integers n ≥0 f((k+1)-1)-1 Vk(P(k) → P(k + 1)) is true, that is, Vk(f(k)=1-k→ ƒ(k+1) = 1 − (k+1)) f(0) = 1-0 =1, which is the value given for f(0) in the definition of f f(k) = 1-k for an arbitrary integer k ≥ 0 f(n)=1-n 1- (k+1) f((k+1)-1)-1 (1-k) - 1 f(k) - 1 Reset Reset
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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Transcribed Image Text:Determine whether each of these proposed definitions is a valid recursive definition of a function ffrom the set of
nonnegative integers to the set of integers. If fis well defined, find a formula for f(n) when n is a nonnegative integer and
prove that your formula is valid.
Prove by mathematical induction that the formula found in the previous problem is valid. First, outline the proof by clicking and
dragging to complete each statement.
Let P(n) be the proposition that
f(n) = 1-n
Basis Step: P(0) states that
f(0) = 1-0 =1, which is the value given for
f(0) in the definition of f
Inductive Step: Assume that
P(n) is true for all integers n ≥0
Show that
f(k)=1-k for an arbitrary integer k≥ 0.
We have completed the basis step and the
inductive step. By mathematical induction, we
know that
Vk(P(k) → P(k+1)) is true, that is,
Vk(f(k) 1-k→ f(k+1)=1-(k+1))
Vk(ƒ(k+1) =
=
IH
=
II
Second, click and drag expressions to fill in the details of showing that VK(P(k)→ P(k+ 1)) is true, thereby completing the induction
step.
(You must provide an answer before moving to the next part.)
(1-k) - 1
f(k) - 1
1- (k+1)
P(n) is true for all integers n ≥0
f((k+1)-1)-1
Vk(P(k) → P(k + 1)) is true, that is,
Vk(f(k)=1-k→ ƒ(k+1) = 1 − (k+1))
f(0) = 1-0 =1, which is the value given for
f(0) in the definition of f
f(k) = 1-k for an arbitrary integer k ≥ 0
f(n)=1-n
1- (k+1)
f((k+1)-1)-1
(1-k) - 1
f(k) - 1
Reset
Reset
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