Computation of partial fraction decomposition
$begingroup$
Suppose the polynomial $Q(x)$ be $$Q(x)=(x-alpha_1)(x-alpha_2) cdots (x-alpha_{n+1})$$, where $alpha_1, alpha_2, cdots , alpha_{n+1}$ are distinct real numbers and $n in mathbb{N}$. Show that if $P(x)$ is a polynomial with degree less than $n+1$ then
$$frac{P(x)}{Q(x)}=sum^{n+1}_{k=1}frac{P(alpha_k)}{Q'(alpha_k)(x-alpha_k)},$$
where $Q'(x)$ is the derivative of the polynomial $Q(x)$.
The generalisation of partial fractions decomposition is strongly related to Lagrange polynomial, should I start proving Lagrange polynomial before the statement above? Or I should have another approach to prove this?
real-analysis derivatives
edited Dec 13 '18 at 14:47
David C. Ullrich
60.9k43994
asked Dec 13 '18 at 14:09
weilam06weilam06
9511
$endgroup$
add a comment |
$begingroup$
Suppose the polynomial $Q(x)$ be $$Q(x)=(x-alpha_1)(x-alpha_2) cdots (x-alpha_{n+1})$$, where $alpha_1, alpha_2, cdots , alpha_{n+1}$ are distinct real numbers and $n in mathbb{N}$. Show that if $P(x)$ is a polynomial with degree less than $n+1$ then
$$frac{P(x)}{Q(x)}=sum^{n+1}_{k=1}frac{P(alpha_k)}{Q'(alpha_k)(x-alpha_k)},$$
where $Q'(x)$ is the derivative of the polynomial $Q(x)$.
The generalisation of partial fractions decomposition is strongly related to Lagrange polynomial, should I start proving Lagrange polynomial before the statement above? Or I should have another approach to prove this?
real-analysis derivatives
edited Dec 13 '18 at 14:47
David C. Ullrich
60.9k43994
asked Dec 13 '18 at 14:09
weilam06weilam06
9511
$endgroup$
add a comment |
$begingroup$
Suppose the polynomial $Q(x)$ be $$Q(x)=(x-alpha_1)(x-alpha_2) cdots (x-alpha_{n+1})$$, where $alpha_1, alpha_2, cdots , alpha_{n+1}$ are distinct real numbers and $n in mathbb{N}$. Show that if $P(x)$ is a polynomial with degree less than $n+1$ then
$$frac{P(x)}{Q(x)}=sum^{n+1}_{k=1}frac{P(alpha_k)}{Q'(alpha_k)(x-alpha_k)},$$
where $Q'(x)$ is the derivative of the polynomial $Q(x)$.
The generalisation of partial fractions decomposition is strongly related to Lagrange polynomial, should I start proving Lagrange polynomial before the statement above? Or I should have another approach to prove this?
real-analysis derivatives
edited Dec 13 '18 at 14:47
David C. Ullrich
60.9k43994
asked Dec 13 '18 at 14:09
weilam06weilam06
9511
$endgroup$
Suppose the polynomial $Q(x)$ be $$Q(x)=(x-alpha_1)(x-alpha_2) cdots (x-alpha_{n+1})$$, where $alpha_1, alpha_2, cdots , alpha_{n+1}$ are distinct real numbers and $n in mathbb{N}$. Show that if $P(x)$ is a polynomial with degree less than $n+1$ then
$$frac{P(x)}{Q(x)}=sum^{n+1}_{k=1}frac{P(alpha_k)}{Q'(alpha_k)(x-alpha_k)},$$
where $Q'(x)$ is the derivative of the polynomial $Q(x)$.
The generalisation of partial fractions decomposition is strongly related to Lagrange polynomial, should I start proving Lagrange polynomial before the statement above? Or I should have another approach to prove this?
real-analysis derivatives
real-analysis derivatives
edited Dec 13 '18 at 14:47
David C. Ullrich
60.9k43994
asked Dec 13 '18 at 14:09
weilam06weilam06
9511
edited Dec 13 '18 at 14:47
David C. Ullrich
60.9k43994
asked Dec 13 '18 at 14:09
weilam06weilam06
9511
edited Dec 13 '18 at 14:47
David C. Ullrich
60.9k43994
edited Dec 13 '18 at 14:47
David C. Ullrich
60.9k43994
edited Dec 13 '18 at 14:47
David C. Ullrich
60.9k43994
60.9k43994
asked Dec 13 '18 at 14:09
weilam06weilam06
9511
asked Dec 13 '18 at 14:09
weilam06weilam06
9511
asked Dec 13 '18 at 14:09
weilam06weilam06
9511
9511
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Two polynomials of degree $n$ are equal if they have the same values at $n+1$ distinct points. Multiply RHS and LHS of your equation by $Q(x)$ and compute the obtained LHS and RHS at any $alpha_j$. If LHS($alpha_j$)=RHS($alpha_j$) are the same for all $j=1,...,n+1$ then the equality is true. Do not forget that
$$
(prod_{j=1}^{n+1}(x-alpha_j))'=sum_{j=1}^{n+1}prod_{ineq j}(x-alpha_i).
$$
answered Dec 13 '18 at 14:30
AAKAAK
365
$endgroup$
$begingroup$
So I have to make $r(x)=P(x)-sum^{n+1}_{k=1}frac{alpha_k}{Q'(alpha_k)(x-alpha_k)} cdot Q(x)=0$ holds true for all $alpha_k$, where $k=1,2, cdots, n+1$. Prove that $r(x)$ with degree $n$ has $n+1$ roots, concludes that $r(x)$ must be a zero polynomial, i.e. $r(x)=0$, so the conclusion follows?
$endgroup$
– weilam06
Dec 13 '18 at 14:55
$begingroup$
Yes! If a polynomial of degree $n$ has at least $n+1$ roots then it is $0$. Just a remark: put $P(alpha_k)$ instead of $alpha_k$ in the numerator in your comment.
$endgroup$
– AAK
Dec 13 '18 at 18:10
$begingroup$
Oh okay, I have mistyped it. Thanks.
$endgroup$
– weilam06
Dec 14 '18 at 8:36
add a comment |
$begingroup$
hint
The decomposition gives
$$frac{P(x)}{Q(x)}=sum_{k=1}^{n+1}frac{A_k}{(x-alpha_k)}$$
with
$$A_k=lim_{xtoalpha_k}frac{P(x)(x-alpha_k)}{Q(x)}$$
and
$$Q'(alpha_k)=lim_{xtoalpha_k}frac{Q(x)-0}{x-alpha_k}.$$
You can take it.
answered Dec 13 '18 at 14:27
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
I don't understand how to relate with MVT.
$endgroup$
– weilam06
Dec 14 '18 at 2:46
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Two polynomials of degree $n$ are equal if they have the same values at $n+1$ distinct points. Multiply RHS and LHS of your equation by $Q(x)$ and compute the obtained LHS and RHS at any $alpha_j$. If LHS($alpha_j$)=RHS($alpha_j$) are the same for all $j=1,...,n+1$ then the equality is true. Do not forget that
$$
(prod_{j=1}^{n+1}(x-alpha_j))'=sum_{j=1}^{n+1}prod_{ineq j}(x-alpha_i).
$$
answered Dec 13 '18 at 14:30
AAKAAK
365
$endgroup$
$begingroup$
So I have to make $r(x)=P(x)-sum^{n+1}_{k=1}frac{alpha_k}{Q'(alpha_k)(x-alpha_k)} cdot Q(x)=0$ holds true for all $alpha_k$, where $k=1,2, cdots, n+1$. Prove that $r(x)$ with degree $n$ has $n+1$ roots, concludes that $r(x)$ must be a zero polynomial, i.e. $r(x)=0$, so the conclusion follows?
$endgroup$
– weilam06
Dec 13 '18 at 14:55
$begingroup$
Yes! If a polynomial of degree $n$ has at least $n+1$ roots then it is $0$. Just a remark: put $P(alpha_k)$ instead of $alpha_k$ in the numerator in your comment.
$endgroup$
– AAK
Dec 13 '18 at 18:10
$begingroup$
Oh okay, I have mistyped it. Thanks.
$endgroup$
– weilam06
Dec 14 '18 at 8:36
add a comment |
$begingroup$
Two polynomials of degree $n$ are equal if they have the same values at $n+1$ distinct points. Multiply RHS and LHS of your equation by $Q(x)$ and compute the obtained LHS and RHS at any $alpha_j$. If LHS($alpha_j$)=RHS($alpha_j$) are the same for all $j=1,...,n+1$ then the equality is true. Do not forget that
$$
(prod_{j=1}^{n+1}(x-alpha_j))'=sum_{j=1}^{n+1}prod_{ineq j}(x-alpha_i).
$$
answered Dec 13 '18 at 14:30
AAKAAK
365
$endgroup$
$begingroup$
So I have to make $r(x)=P(x)-sum^{n+1}_{k=1}frac{alpha_k}{Q'(alpha_k)(x-alpha_k)} cdot Q(x)=0$ holds true for all $alpha_k$, where $k=1,2, cdots, n+1$. Prove that $r(x)$ with degree $n$ has $n+1$ roots, concludes that $r(x)$ must be a zero polynomial, i.e. $r(x)=0$, so the conclusion follows?
$endgroup$
– weilam06
Dec 13 '18 at 14:55
$begingroup$
Yes! If a polynomial of degree $n$ has at least $n+1$ roots then it is $0$. Just a remark: put $P(alpha_k)$ instead of $alpha_k$ in the numerator in your comment.
$endgroup$
– AAK
Dec 13 '18 at 18:10
$begingroup$
Oh okay, I have mistyped it. Thanks.
$endgroup$
– weilam06
Dec 14 '18 at 8:36
add a comment |
$begingroup$
Two polynomials of degree $n$ are equal if they have the same values at $n+1$ distinct points. Multiply RHS and LHS of your equation by $Q(x)$ and compute the obtained LHS and RHS at any $alpha_j$. If LHS($alpha_j$)=RHS($alpha_j$) are the same for all $j=1,...,n+1$ then the equality is true. Do not forget that
$$
(prod_{j=1}^{n+1}(x-alpha_j))'=sum_{j=1}^{n+1}prod_{ineq j}(x-alpha_i).
$$
answered Dec 13 '18 at 14:30
AAKAAK
365
$endgroup$
Two polynomials of degree $n$ are equal if they have the same values at $n+1$ distinct points. Multiply RHS and LHS of your equation by $Q(x)$ and compute the obtained LHS and RHS at any $alpha_j$. If LHS($alpha_j$)=RHS($alpha_j$) are the same for all $j=1,...,n+1$ then the equality is true. Do not forget that
$$
(prod_{j=1}^{n+1}(x-alpha_j))'=sum_{j=1}^{n+1}prod_{ineq j}(x-alpha_i).
$$
answered Dec 13 '18 at 14:30
AAKAAK
365
answered Dec 13 '18 at 14:30
AAKAAK
365
answered Dec 13 '18 at 14:30
AAKAAK
365
answered Dec 13 '18 at 14:30
AAKAAK
365
365
$begingroup$
So I have to make $r(x)=P(x)-sum^{n+1}_{k=1}frac{alpha_k}{Q'(alpha_k)(x-alpha_k)} cdot Q(x)=0$ holds true for all $alpha_k$, where $k=1,2, cdots, n+1$. Prove that $r(x)$ with degree $n$ has $n+1$ roots, concludes that $r(x)$ must be a zero polynomial, i.e. $r(x)=0$, so the conclusion follows?
$endgroup$
– weilam06
Dec 13 '18 at 14:55
$begingroup$
Yes! If a polynomial of degree $n$ has at least $n+1$ roots then it is $0$. Just a remark: put $P(alpha_k)$ instead of $alpha_k$ in the numerator in your comment.
$endgroup$
– AAK
Dec 13 '18 at 18:10
$begingroup$
Oh okay, I have mistyped it. Thanks.
$endgroup$
– weilam06
Dec 14 '18 at 8:36
add a comment |
$begingroup$
So I have to make $r(x)=P(x)-sum^{n+1}_{k=1}frac{alpha_k}{Q'(alpha_k)(x-alpha_k)} cdot Q(x)=0$ holds true for all $alpha_k$, where $k=1,2, cdots, n+1$. Prove that $r(x)$ with degree $n$ has $n+1$ roots, concludes that $r(x)$ must be a zero polynomial, i.e. $r(x)=0$, so the conclusion follows?
$endgroup$
– weilam06
Dec 13 '18 at 14:55
$begingroup$
Yes! If a polynomial of degree $n$ has at least $n+1$ roots then it is $0$. Just a remark: put $P(alpha_k)$ instead of $alpha_k$ in the numerator in your comment.
$endgroup$
– AAK
Dec 13 '18 at 18:10
$begingroup$
Oh okay, I have mistyped it. Thanks.
$endgroup$
– weilam06
Dec 14 '18 at 8:36
$begingroup$
So I have to make $r(x)=P(x)-sum^{n+1}_{k=1}frac{alpha_k}{Q'(alpha_k)(x-alpha_k)} cdot Q(x)=0$ holds true for all $alpha_k$, where $k=1,2, cdots, n+1$. Prove that $r(x)$ with degree $n$ has $n+1$ roots, concludes that $r(x)$ must be a zero polynomial, i.e. $r(x)=0$, so the conclusion follows?
$endgroup$
– weilam06
Dec 13 '18 at 14:55
$begingroup$
So I have to make $r(x)=P(x)-sum^{n+1}_{k=1}frac{alpha_k}{Q'(alpha_k)(x-alpha_k)} cdot Q(x)=0$ holds true for all $alpha_k$, where $k=1,2, cdots, n+1$. Prove that $r(x)$ with degree $n$ has $n+1$ roots, concludes that $r(x)$ must be a zero polynomial, i.e. $r(x)=0$, so the conclusion follows?
$endgroup$
– weilam06
Dec 13 '18 at 14:55
$begingroup$
Yes! If a polynomial of degree $n$ has at least $n+1$ roots then it is $0$. Just a remark: put $P(alpha_k)$ instead of $alpha_k$ in the numerator in your comment.
$endgroup$
– AAK
Dec 13 '18 at 18:10
$begingroup$
Yes! If a polynomial of degree $n$ has at least $n+1$ roots then it is $0$. Just a remark: put $P(alpha_k)$ instead of $alpha_k$ in the numerator in your comment.
$endgroup$
– AAK
Dec 13 '18 at 18:10
$begingroup$
Oh okay, I have mistyped it. Thanks.
$endgroup$
– weilam06
Dec 14 '18 at 8:36
$begingroup$
Oh okay, I have mistyped it. Thanks.
$endgroup$
– weilam06
Dec 14 '18 at 8:36
add a comment |
$begingroup$
hint
The decomposition gives
$$frac{P(x)}{Q(x)}=sum_{k=1}^{n+1}frac{A_k}{(x-alpha_k)}$$
with
$$A_k=lim_{xtoalpha_k}frac{P(x)(x-alpha_k)}{Q(x)}$$
and
$$Q'(alpha_k)=lim_{xtoalpha_k}frac{Q(x)-0}{x-alpha_k}.$$
You can take it.
answered Dec 13 '18 at 14:27
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
I don't understand how to relate with MVT.
$endgroup$
– weilam06
Dec 14 '18 at 2:46
add a comment |
$begingroup$
hint
The decomposition gives
$$frac{P(x)}{Q(x)}=sum_{k=1}^{n+1}frac{A_k}{(x-alpha_k)}$$
with
$$A_k=lim_{xtoalpha_k}frac{P(x)(x-alpha_k)}{Q(x)}$$
and
$$Q'(alpha_k)=lim_{xtoalpha_k}frac{Q(x)-0}{x-alpha_k}.$$
You can take it.
answered Dec 13 '18 at 14:27
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
I don't understand how to relate with MVT.
$endgroup$
– weilam06
Dec 14 '18 at 2:46
add a comment |
$begingroup$
hint
The decomposition gives
$$frac{P(x)}{Q(x)}=sum_{k=1}^{n+1}frac{A_k}{(x-alpha_k)}$$
with
$$A_k=lim_{xtoalpha_k}frac{P(x)(x-alpha_k)}{Q(x)}$$
and
$$Q'(alpha_k)=lim_{xtoalpha_k}frac{Q(x)-0}{x-alpha_k}.$$
You can take it.
answered Dec 13 '18 at 14:27
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
hint
The decomposition gives
$$frac{P(x)}{Q(x)}=sum_{k=1}^{n+1}frac{A_k}{(x-alpha_k)}$$
with
$$A_k=lim_{xtoalpha_k}frac{P(x)(x-alpha_k)}{Q(x)}$$
and
$$Q'(alpha_k)=lim_{xtoalpha_k}frac{Q(x)-0}{x-alpha_k}.$$
You can take it.
answered Dec 13 '18 at 14:27
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 13 '18 at 14:27
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 13 '18 at 14:27
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 13 '18 at 14:27
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
$begingroup$
I don't understand how to relate with MVT.
$endgroup$
– weilam06
Dec 14 '18 at 2:46
add a comment |
$begingroup$
I don't understand how to relate with MVT.
$endgroup$
– weilam06
Dec 14 '18 at 2:46
$begingroup$
I don't understand how to relate with MVT.
$endgroup$
– weilam06
Dec 14 '18 at 2:46
$begingroup$
I don't understand how to relate with MVT.
$endgroup$
– weilam06
Dec 14 '18 at 2:46
add a comment |
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