• If f and g are completely multiplicative arithmetic functions, then f *g must also be completely multiplicative. Proof: : дcd(m, п) .. Any divisor of mn is the form d = d,d2 where d1 | m and d, | n * m, n has no shared divisor .. f * g(mn) = Ea edmn (d)g() = Edilm Edzln F(d1d2)g() Σ Σm f(di)f(da)g(Η )g(H) . Σmf (d1)g(H) Σ.u f (d%) g ( H) :: (f * g)(m)(f * g)(n) • If f and g are arithmetic functions and f * g is completely multiplicative, then dı|m 2|n both f and g must be completely multiplicative. • Convolution of arithmetic functions is associative: (f * g) * h = f * (g * h) for arithmetic functions f, g and h.
• If f and g are completely multiplicative arithmetic functions, then f *g must also be completely multiplicative. Proof: : дcd(m, п) .. Any divisor of mn is the form d = d,d2 where d1 | m and d, | n * m, n has no shared divisor .. f * g(mn) = Ea edmn (d)g() = Edilm Edzln F(d1d2)g() Σ Σm f(di)f(da)g(Η )g(H) . Σmf (d1)g(H) Σ.u f (d%) g ( H) :: (f * g)(m)(f * g)(n) • If f and g are arithmetic functions and f * g is completely multiplicative, then dı|m 2|n both f and g must be completely multiplicative. • Convolution of arithmetic functions is associative: (f * g) * h = f * (g * h) for arithmetic functions f, g and h.
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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Question
Please do the second and third one thanks!
![1. Prove or provide a counterexample to each of the following statements:
• If f and g are completely multiplicative arithmetic functions, then f *g must also
be completely multiplicative.
Proof:
* gcd(m, n)
. Any divisor of mn is the form d = d,d, where d | m and d, | n
* m, n has no shared divisor
.. f * g(mn) = Eamn f (d)g() = di|m Edz]n f(d1d2)g()
= 1
тп
: Σm Σn f(di)f(d)g(Ή)g(X)
. Σf(d1)g(H) Σιμ f(d)g( )
dı|m
d2|n
:: (f * g)(m)(f * g)(n)
• If f and g are arithmetic functions and f * g is completely multiplicative, then
both f and g must be completely multiplicative.
• Convolution of arithmetic functions is associative: (f * g) * h = f * (g * h) for
arithmetic functions f, g and h.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd86ff4db-1734-44b4-a0ef-524093ba313a%2Fbca305fc-083f-418f-b29c-8de1e07c00e8%2Fq9nvc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. Prove or provide a counterexample to each of the following statements:
• If f and g are completely multiplicative arithmetic functions, then f *g must also
be completely multiplicative.
Proof:
* gcd(m, n)
. Any divisor of mn is the form d = d,d, where d | m and d, | n
* m, n has no shared divisor
.. f * g(mn) = Eamn f (d)g() = di|m Edz]n f(d1d2)g()
= 1
тп
: Σm Σn f(di)f(d)g(Ή)g(X)
. Σf(d1)g(H) Σιμ f(d)g( )
dı|m
d2|n
:: (f * g)(m)(f * g)(n)
• If f and g are arithmetic functions and f * g is completely multiplicative, then
both f and g must be completely multiplicative.
• Convolution of arithmetic functions is associative: (f * g) * h = f * (g * h) for
arithmetic functions f, g and h.
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