Mixing
Showing the Composition of Two Polynomials is a Polynomial and the Composition of Two Rational Functions is a Rational Function
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This seems very obvious and I am having a bit of trouble producing a formal proof.
sketch proof that the composition of two polynomials is a polynomial
Let $$p(z_1)=a_nz^n_1+a_n-1z^n-1_1+...+a_1z_1+a_0 \ q(z_2)=b_nz^n_2+b_n-1z^n-1_2+...+b_1z_2+b_0$$ be two complex polynomials of degree $n$ where $a_n,..,a_0inmathbbC$ and $b_n,..,b_oinmathbbC$.
Now,
beginalign
(pcirc q)(z_2)&=p(q(z_2)) text(by definition)\
&=a_n(q(z_2))^n+a_n-1(q(z_2))^n-1+...+a_1(q(z_2))+a_0
endalign
which is clearly a complex polynomial of degree $n^2$.
sketch proof that the composition of two rational functions is a rational function
A rational function is a quotient of polynomials.
Let $$a(z_1)=fracp(z_1)q(z_1), b(z_2)=fracp(z_2)q(z_2)$$
Now,
beginalign
(acirc b)(z_2)&=a(b(z_2)) text(by definition) \
&=fracpleft(fracp(z_2)q(z_2)right)qleft(fracp(z_2)q(z_2)right) \
&=fraca_nleft(fracp(z_2)q(z_2)right)^n+a_n-1left(fracp(z_2)q(z_2)right)^n-1+...+a_1left(fracp(z_2)q(z_2)right)+a_0b_nleft(fracp(z_2)q(z_2)right)^n+b_n-1left(fracp(z_2)q(z_2)right)^n-1+...+b_1left(fracp(z_2)q(z_2)right)+b_0 \
endalign
Notice that $left(fracp(z_2)q(z_2)right)^i (i=n, n-1,..,0)$ is a polynomial as
$$(fcirc g)(z_2)=f(g(z_2))=left(fracp(z_2)q(z_2)right)^i$$
where $$f(x)=x^i, g(z_2)=left(fracp(z_2)q(z_2)right)$$ are both polynomials.
Hence $(acirc b)(z_2)$ is a rational function as it is the quotient of polynomials.
complex-analysis functions proof-verification proof-writing
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This seems very obvious and I am having a bit of trouble producing a formal proof.
sketch proof that the composition of two polynomials is a polynomial
Let $$p(z_1)=a_nz^n_1+a_n-1z^n-1_1+...+a_1z_1+a_0 \ q(z_2)=b_nz^n_2+b_n-1z^n-1_2+...+b_1z_2+b_0$$ be two complex polynomials of degree $n$ where $a_n,..,a_0inmathbbC$ and $b_n,..,b_oinmathbbC$.
Now,
beginalign
(pcirc q)(z_2)&=p(q(z_2)) text(by definition)\
&=a_n(q(z_2))^n+a_n-1(q(z_2))^n-1+...+a_1(q(z_2))+a_0
endalign
which is clearly a complex polynomial of degree $n^2$.
sketch proof that the composition of two rational functions is a rational function
A rational function is a quotient of polynomials.
Let $$a(z_1)=fracp(z_1)q(z_1), b(z_2)=fracp(z_2)q(z_2)$$
Now,
beginalign
(acirc b)(z_2)&=a(b(z_2)) text(by definition) \
&=fracpleft(fracp(z_2)q(z_2)right)qleft(fracp(z_2)q(z_2)right) \
&=fraca_nleft(fracp(z_2)q(z_2)right)^n+a_n-1left(fracp(z_2)q(z_2)right)^n-1+...+a_1left(fracp(z_2)q(z_2)right)+a_0b_nleft(fracp(z_2)q(z_2)right)^n+b_n-1left(fracp(z_2)q(z_2)right)^n-1+...+b_1left(fracp(z_2)q(z_2)right)+b_0 \
endalign
Notice that $left(fracp(z_2)q(z_2)right)^i (i=n, n-1,..,0)$ is a polynomial as
$$(fcirc g)(z_2)=f(g(z_2))=left(fracp(z_2)q(z_2)right)^i$$
where $$f(x)=x^i, g(z_2)=left(fracp(z_2)q(z_2)right)$$ are both polynomials.
Hence $(acirc b)(z_2)$ is a rational function as it is the quotient of polynomials.
complex-analysis functions proof-verification proof-writing
asked 2 days ago
Bell
54312
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This seems very obvious and I am having a bit of trouble producing a formal proof.
sketch proof that the composition of two polynomials is a polynomial
Let $$p(z_1)=a_nz^n_1+a_n-1z^n-1_1+...+a_1z_1+a_0 \ q(z_2)=b_nz^n_2+b_n-1z^n-1_2+...+b_1z_2+b_0$$ be two complex polynomials of degree $n$ where $a_n,..,a_0inmathbbC$ and $b_n,..,b_oinmathbbC$.
Now,
beginalign
(pcirc q)(z_2)&=p(q(z_2)) text(by definition)\
&=a_n(q(z_2))^n+a_n-1(q(z_2))^n-1+...+a_1(q(z_2))+a_0
endalign
which is clearly a complex polynomial of degree $n^2$.
sketch proof that the composition of two rational functions is a rational function
A rational function is a quotient of polynomials.
Let $$a(z_1)=fracp(z_1)q(z_1), b(z_2)=fracp(z_2)q(z_2)$$
Now,
beginalign
(acirc b)(z_2)&=a(b(z_2)) text(by definition) \
&=fracpleft(fracp(z_2)q(z_2)right)qleft(fracp(z_2)q(z_2)right) \
&=fraca_nleft(fracp(z_2)q(z_2)right)^n+a_n-1left(fracp(z_2)q(z_2)right)^n-1+...+a_1left(fracp(z_2)q(z_2)right)+a_0b_nleft(fracp(z_2)q(z_2)right)^n+b_n-1left(fracp(z_2)q(z_2)right)^n-1+...+b_1left(fracp(z_2)q(z_2)right)+b_0 \
endalign
Notice that $left(fracp(z_2)q(z_2)right)^i (i=n, n-1,..,0)$ is a polynomial as
$$(fcirc g)(z_2)=f(g(z_2))=left(fracp(z_2)q(z_2)right)^i$$
where $$f(x)=x^i, g(z_2)=left(fracp(z_2)q(z_2)right)$$ are both polynomials.
Hence $(acirc b)(z_2)$ is a rational function as it is the quotient of polynomials.
complex-analysis functions proof-verification proof-writing
asked 2 days ago
Bell
54312
This seems very obvious and I am having a bit of trouble producing a formal proof.
sketch proof that the composition of two polynomials is a polynomial
Let $$p(z_1)=a_nz^n_1+a_n-1z^n-1_1+...+a_1z_1+a_0 \ q(z_2)=b_nz^n_2+b_n-1z^n-1_2+...+b_1z_2+b_0$$ be two complex polynomials of degree $n$ where $a_n,..,a_0inmathbbC$ and $b_n,..,b_oinmathbbC$.
Now,
beginalign
(pcirc q)(z_2)&=p(q(z_2)) text(by definition)\
&=a_n(q(z_2))^n+a_n-1(q(z_2))^n-1+...+a_1(q(z_2))+a_0
endalign
which is clearly a complex polynomial of degree $n^2$.
sketch proof that the composition of two rational functions is a rational function
A rational function is a quotient of polynomials.
Let $$a(z_1)=fracp(z_1)q(z_1), b(z_2)=fracp(z_2)q(z_2)$$
Now,
beginalign
(acirc b)(z_2)&=a(b(z_2)) text(by definition) \
&=fracpleft(fracp(z_2)q(z_2)right)qleft(fracp(z_2)q(z_2)right) \
&=fraca_nleft(fracp(z_2)q(z_2)right)^n+a_n-1left(fracp(z_2)q(z_2)right)^n-1+...+a_1left(fracp(z_2)q(z_2)right)+a_0b_nleft(fracp(z_2)q(z_2)right)^n+b_n-1left(fracp(z_2)q(z_2)right)^n-1+...+b_1left(fracp(z_2)q(z_2)right)+b_0 \
endalign
Notice that $left(fracp(z_2)q(z_2)right)^i (i=n, n-1,..,0)$ is a polynomial as
$$(fcirc g)(z_2)=f(g(z_2))=left(fracp(z_2)q(z_2)right)^i$$
where $$f(x)=x^i, g(z_2)=left(fracp(z_2)q(z_2)right)$$ are both polynomials.
Hence $(acirc b)(z_2)$ is a rational function as it is the quotient of polynomials.
complex-analysis functions proof-verification proof-writing
asked 2 days ago
Bell
54312
asked 2 days ago
Bell
54312
asked 2 days ago
Bell
54312
asked 2 days ago
Bell
54312
54312
add a comment |Â
add a comment |Â
1 Answer
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Remark for your proof of composition of polynomials is a polynomial:
- perhaps you should work with two arbitrary polynomials of degree $m$ and $n$ to have generality.
Remark for your proof of composition of composition rational functions is a rational function:
you wrote $a(z_1)=fracp(z_1)q(z_1)$ and $b(z_2) = fracp(z_2)q(z_2)$ which means $a$ and $b$ seems to be the same function.
$left( fracp(z_1)q(z_2)right)^i$ is a rational function rather than a polynomial.
Guide for proof of composition of polynomials is a polynomial:
- First prove that products of two polynomials is a polynomial. Once you can do that, we have that suppose $p$ is a polynomial, then $p(x)^i$ is a polynomials by mathematical induction.
- Prove that the set of polynomials is closed under scalar multiplication.
- Prove that the set of polynomials is closed under addition.
- With those lemmas (tools), I believe now you can prove that composition of polynomials is a polynomials. (remember to use arbitrary polynomials of degree $m$ and degree $n$.)
Guide for proof of composition of rational functions is a rational function:
- First prove that products of two rational function is a rational function. Once you can do that, we have that suppose $h$ is a rational function, then $h(x)^i$ is a rational function by mathematical induction.
- Prove that the set of rational functions is closed under scalar multiplication.
- Prove that the set of rational function is closed under addition.
- Prove that the set of rational function is closed under division.
- With those lemmas (tools), I believe now you can prove that composition of rational functions is a rational function. Let $a(z)= fracp(z)q(z)$ and $b(z) = fracr(z)s(z)$ and use those tools that you have verified.
answered 10 hours ago
Siong Thye Goh
76.5k124594
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Remark for your proof of composition of polynomials is a polynomial:
- perhaps you should work with two arbitrary polynomials of degree $m$ and $n$ to have generality.
Remark for your proof of composition of composition rational functions is a rational function:
you wrote $a(z_1)=fracp(z_1)q(z_1)$ and $b(z_2) = fracp(z_2)q(z_2)$ which means $a$ and $b$ seems to be the same function.
$left( fracp(z_1)q(z_2)right)^i$ is a rational function rather than a polynomial.
Guide for proof of composition of polynomials is a polynomial:
- First prove that products of two polynomials is a polynomial. Once you can do that, we have that suppose $p$ is a polynomial, then $p(x)^i$ is a polynomials by mathematical induction.
- Prove that the set of polynomials is closed under scalar multiplication.
- Prove that the set of polynomials is closed under addition.
- With those lemmas (tools), I believe now you can prove that composition of polynomials is a polynomials. (remember to use arbitrary polynomials of degree $m$ and degree $n$.)
Guide for proof of composition of rational functions is a rational function:
- First prove that products of two rational function is a rational function. Once you can do that, we have that suppose $h$ is a rational function, then $h(x)^i$ is a rational function by mathematical induction.
- Prove that the set of rational functions is closed under scalar multiplication.
- Prove that the set of rational function is closed under addition.
- Prove that the set of rational function is closed under division.
- With those lemmas (tools), I believe now you can prove that composition of rational functions is a rational function. Let $a(z)= fracp(z)q(z)$ and $b(z) = fracr(z)s(z)$ and use those tools that you have verified.
answered 10 hours ago
Siong Thye Goh
76.5k124594
add a comment |Â
up vote
1
down vote
accepted
Remark for your proof of composition of polynomials is a polynomial:
- perhaps you should work with two arbitrary polynomials of degree $m$ and $n$ to have generality.
Remark for your proof of composition of composition rational functions is a rational function:
you wrote $a(z_1)=fracp(z_1)q(z_1)$ and $b(z_2) = fracp(z_2)q(z_2)$ which means $a$ and $b$ seems to be the same function.
$left( fracp(z_1)q(z_2)right)^i$ is a rational function rather than a polynomial.
Guide for proof of composition of polynomials is a polynomial:
- First prove that products of two polynomials is a polynomial. Once you can do that, we have that suppose $p$ is a polynomial, then $p(x)^i$ is a polynomials by mathematical induction.
- Prove that the set of polynomials is closed under scalar multiplication.
- Prove that the set of polynomials is closed under addition.
- With those lemmas (tools), I believe now you can prove that composition of polynomials is a polynomials. (remember to use arbitrary polynomials of degree $m$ and degree $n$.)
Guide for proof of composition of rational functions is a rational function:
- First prove that products of two rational function is a rational function. Once you can do that, we have that suppose $h$ is a rational function, then $h(x)^i$ is a rational function by mathematical induction.
- Prove that the set of rational functions is closed under scalar multiplication.
- Prove that the set of rational function is closed under addition.
- Prove that the set of rational function is closed under division.
- With those lemmas (tools), I believe now you can prove that composition of rational functions is a rational function. Let $a(z)= fracp(z)q(z)$ and $b(z) = fracr(z)s(z)$ and use those tools that you have verified.
answered 10 hours ago
Siong Thye Goh
76.5k124594
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Remark for your proof of composition of polynomials is a polynomial:
- perhaps you should work with two arbitrary polynomials of degree $m$ and $n$ to have generality.
Remark for your proof of composition of composition rational functions is a rational function:
you wrote $a(z_1)=fracp(z_1)q(z_1)$ and $b(z_2) = fracp(z_2)q(z_2)$ which means $a$ and $b$ seems to be the same function.
$left( fracp(z_1)q(z_2)right)^i$ is a rational function rather than a polynomial.
Guide for proof of composition of polynomials is a polynomial:
- First prove that products of two polynomials is a polynomial. Once you can do that, we have that suppose $p$ is a polynomial, then $p(x)^i$ is a polynomials by mathematical induction.
- Prove that the set of polynomials is closed under scalar multiplication.
- Prove that the set of polynomials is closed under addition.
- With those lemmas (tools), I believe now you can prove that composition of polynomials is a polynomials. (remember to use arbitrary polynomials of degree $m$ and degree $n$.)
Guide for proof of composition of rational functions is a rational function:
- First prove that products of two rational function is a rational function. Once you can do that, we have that suppose $h$ is a rational function, then $h(x)^i$ is a rational function by mathematical induction.
- Prove that the set of rational functions is closed under scalar multiplication.
- Prove that the set of rational function is closed under addition.
- Prove that the set of rational function is closed under division.
- With those lemmas (tools), I believe now you can prove that composition of rational functions is a rational function. Let $a(z)= fracp(z)q(z)$ and $b(z) = fracr(z)s(z)$ and use those tools that you have verified.
answered 10 hours ago
Siong Thye Goh
76.5k124594
Remark for your proof of composition of polynomials is a polynomial:
- perhaps you should work with two arbitrary polynomials of degree $m$ and $n$ to have generality.
Remark for your proof of composition of composition rational functions is a rational function:
you wrote $a(z_1)=fracp(z_1)q(z_1)$ and $b(z_2) = fracp(z_2)q(z_2)$ which means $a$ and $b$ seems to be the same function.
$left( fracp(z_1)q(z_2)right)^i$ is a rational function rather than a polynomial.
Guide for proof of composition of polynomials is a polynomial:
- First prove that products of two polynomials is a polynomial. Once you can do that, we have that suppose $p$ is a polynomial, then $p(x)^i$ is a polynomials by mathematical induction.
- Prove that the set of polynomials is closed under scalar multiplication.
- Prove that the set of polynomials is closed under addition.
- With those lemmas (tools), I believe now you can prove that composition of polynomials is a polynomials. (remember to use arbitrary polynomials of degree $m$ and degree $n$.)
Guide for proof of composition of rational functions is a rational function:
- First prove that products of two rational function is a rational function. Once you can do that, we have that suppose $h$ is a rational function, then $h(x)^i$ is a rational function by mathematical induction.
- Prove that the set of rational functions is closed under scalar multiplication.
- Prove that the set of rational function is closed under addition.
- Prove that the set of rational function is closed under division.
- With those lemmas (tools), I believe now you can prove that composition of rational functions is a rational function. Let $a(z)= fracp(z)q(z)$ and $b(z) = fracr(z)s(z)$ and use those tools that you have verified.
answered 10 hours ago
Siong Thye Goh
76.5k124594
answered 10 hours ago
Siong Thye Goh
76.5k124594
answered 10 hours ago
Siong Thye Goh
76.5k124594
answered 10 hours ago
Siong Thye Goh
76.5k124594
76.5k124594
add a comment |Â
add a comment |Â
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