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Multiplying two polynomials: explanation of the general formula for the coefficients
Getting a bound on the coefficients of the factor polynomialFinding the sum of the coefficients of polynomial of degree 21Determining if any of these three are an ideal of $mathbbR[x]$In $R[x]$, $f=g iff f(x)=g(x), forall x in R$Deriving $x^n-1=(x-1)(x^n-1+x^n-2+…+x+1)$Is sum of two stable polynomials still stable?Irreducibility criteria for polynomials in $mathbbZ[x]$Common zeros of two univariate polynomials over $mathbbC(y)$Help Understanding Proof of Eisenstein's CriterionHow determine relationship between two $n$ degree polynomials roots
$begingroup$
If $$f(x)=a_nx^n+...+a_0$$ and $$g(x)=b_mx^m+…+b_0$$ then $$f(x)cdot g(x)=c_m+nx^m+n+...+c_0$$ where $c_k=sum_r+s=ka_rb_s,quad k=0,....,m+n$
I know that the degree $0$ and $(m+n)$ exists, and I also understand the Formula for those cases.
Now if I pick a degree $k$ in between $0$ and $(m+n)$. Then if I always take a summand of the first factor $f(x)$ i. e. $, a_i$, then $i$ cannot be bigger than $k$ because I would have to multiply it with a monomonial with at least degree $0$.
Those are my thoughts so far; how can I continue this chain of thoughts in order to prove the Formula for the coefficients, i. e. why does it have to be this Formula?
sequences-and-series polynomials cauchy-product
edited 17 hours ago
Martin Hansen
19113
asked yesterday
New2MathNew2Math
8912
$endgroup$
add a comment |
$begingroup$
If $$f(x)=a_nx^n+...+a_0$$ and $$g(x)=b_mx^m+…+b_0$$ then $$f(x)cdot g(x)=c_m+nx^m+n+...+c_0$$ where $c_k=sum_r+s=ka_rb_s,quad k=0,....,m+n$
I know that the degree $0$ and $(m+n)$ exists, and I also understand the Formula for those cases.
Now if I pick a degree $k$ in between $0$ and $(m+n)$. Then if I always take a summand of the first factor $f(x)$ i. e. $, a_i$, then $i$ cannot be bigger than $k$ because I would have to multiply it with a monomonial with at least degree $0$.
Those are my thoughts so far; how can I continue this chain of thoughts in order to prove the Formula for the coefficients, i. e. why does it have to be this Formula?
sequences-and-series polynomials cauchy-product
edited 17 hours ago
Martin Hansen
19113
asked yesterday
New2MathNew2Math
8912
$endgroup$
$begingroup$
Do you mean fg(x) which is composite functions ? I suspect from the title you mean f(x)g(x). If so, looking up "Cauchy Product" may help.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
@MartinHansen I have tried to prove the Formula with induction First if $m=0,n=0$ the Formula holds. Then Setting $m>0$ and $n=0$ The Formula still holds and then I make an iduction over $n$ and in the inductionstep after using the assocaitive and distributive law I have got the term $a_n+1b_mx^n+1+m+(c_n+1+m-1+a_n+1b_m-1)x^n+1+m-1+...+(c_n+1+a_n+1b_0)x^n+1+c_nx^n+...c_0$. Can you tell me how I have to Phrase the inductionhypothesis to get the desired result?
$endgroup$
– New2Math
19 hours ago
$begingroup$
I think it should be possible to get that to work although it's probably hiding what's going on, rather than illuminating the mechanism at work. However, your comment got me thinking and I've posted an answer below that I think is more straight forward to understand. Hope you like it !
$endgroup$
– Martin Hansen
18 hours ago
add a comment |
$begingroup$
If $$f(x)=a_nx^n+...+a_0$$ and $$g(x)=b_mx^m+…+b_0$$ then $$f(x)cdot g(x)=c_m+nx^m+n+...+c_0$$ where $c_k=sum_r+s=ka_rb_s,quad k=0,....,m+n$
I know that the degree $0$ and $(m+n)$ exists, and I also understand the Formula for those cases.
Now if I pick a degree $k$ in between $0$ and $(m+n)$. Then if I always take a summand of the first factor $f(x)$ i. e. $, a_i$, then $i$ cannot be bigger than $k$ because I would have to multiply it with a monomonial with at least degree $0$.
Those are my thoughts so far; how can I continue this chain of thoughts in order to prove the Formula for the coefficients, i. e. why does it have to be this Formula?
sequences-and-series polynomials cauchy-product
edited 17 hours ago
Martin Hansen
19113
asked yesterday
New2MathNew2Math
8912
$endgroup$
If $$f(x)=a_nx^n+...+a_0$$ and $$g(x)=b_mx^m+…+b_0$$ then $$f(x)cdot g(x)=c_m+nx^m+n+...+c_0$$ where $c_k=sum_r+s=ka_rb_s,quad k=0,....,m+n$
I know that the degree $0$ and $(m+n)$ exists, and I also understand the Formula for those cases.
Now if I pick a degree $k$ in between $0$ and $(m+n)$. Then if I always take a summand of the first factor $f(x)$ i. e. $, a_i$, then $i$ cannot be bigger than $k$ because I would have to multiply it with a monomonial with at least degree $0$.
Those are my thoughts so far; how can I continue this chain of thoughts in order to prove the Formula for the coefficients, i. e. why does it have to be this Formula?
sequences-and-series polynomials cauchy-product
sequences-and-series polynomials cauchy-product
edited 17 hours ago
Martin Hansen
19113
asked yesterday
New2MathNew2Math
8912
edited 17 hours ago
Martin Hansen
19113
asked yesterday
New2MathNew2Math
8912
edited 17 hours ago
Martin Hansen
19113
edited 17 hours ago
Martin Hansen
19113
edited 17 hours ago
Martin Hansen
19113
19113
asked yesterday
New2MathNew2Math
8912
asked yesterday
New2MathNew2Math
8912
asked yesterday
New2MathNew2Math
8912
8912
$begingroup$
Do you mean fg(x) which is composite functions ? I suspect from the title you mean f(x)g(x). If so, looking up "Cauchy Product" may help.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
@MartinHansen I have tried to prove the Formula with induction First if $m=0,n=0$ the Formula holds. Then Setting $m>0$ and $n=0$ The Formula still holds and then I make an iduction over $n$ and in the inductionstep after using the assocaitive and distributive law I have got the term $a_n+1b_mx^n+1+m+(c_n+1+m-1+a_n+1b_m-1)x^n+1+m-1+...+(c_n+1+a_n+1b_0)x^n+1+c_nx^n+...c_0$. Can you tell me how I have to Phrase the inductionhypothesis to get the desired result?
$endgroup$
– New2Math
19 hours ago
$begingroup$
I think it should be possible to get that to work although it's probably hiding what's going on, rather than illuminating the mechanism at work. However, your comment got me thinking and I've posted an answer below that I think is more straight forward to understand. Hope you like it !
$endgroup$
– Martin Hansen
18 hours ago
add a comment |
$begingroup$
Do you mean fg(x) which is composite functions ? I suspect from the title you mean f(x)g(x). If so, looking up "Cauchy Product" may help.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
@MartinHansen I have tried to prove the Formula with induction First if $m=0,n=0$ the Formula holds. Then Setting $m>0$ and $n=0$ The Formula still holds and then I make an iduction over $n$ and in the inductionstep after using the assocaitive and distributive law I have got the term $a_n+1b_mx^n+1+m+(c_n+1+m-1+a_n+1b_m-1)x^n+1+m-1+...+(c_n+1+a_n+1b_0)x^n+1+c_nx^n+...c_0$. Can you tell me how I have to Phrase the inductionhypothesis to get the desired result?
$endgroup$
– New2Math
19 hours ago
$begingroup$
I think it should be possible to get that to work although it's probably hiding what's going on, rather than illuminating the mechanism at work. However, your comment got me thinking and I've posted an answer below that I think is more straight forward to understand. Hope you like it !
$endgroup$
– Martin Hansen
18 hours ago
$begingroup$
Do you mean fg(x) which is composite functions ? I suspect from the title you mean f(x)g(x). If so, looking up "Cauchy Product" may help.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
Do you mean fg(x) which is composite functions ? I suspect from the title you mean f(x)g(x). If so, looking up "Cauchy Product" may help.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
@MartinHansen I have tried to prove the Formula with induction First if $m=0,n=0$ the Formula holds. Then Setting $m>0$ and $n=0$ The Formula still holds and then I make an iduction over $n$ and in the inductionstep after using the assocaitive and distributive law I have got the term $a_n+1b_mx^n+1+m+(c_n+1+m-1+a_n+1b_m-1)x^n+1+m-1+...+(c_n+1+a_n+1b_0)x^n+1+c_nx^n+...c_0$. Can you tell me how I have to Phrase the inductionhypothesis to get the desired result?
$endgroup$
– New2Math
19 hours ago
$begingroup$
@MartinHansen I have tried to prove the Formula with induction First if $m=0,n=0$ the Formula holds. Then Setting $m>0$ and $n=0$ The Formula still holds and then I make an iduction over $n$ and in the inductionstep after using the assocaitive and distributive law I have got the term $a_n+1b_mx^n+1+m+(c_n+1+m-1+a_n+1b_m-1)x^n+1+m-1+...+(c_n+1+a_n+1b_0)x^n+1+c_nx^n+...c_0$. Can you tell me how I have to Phrase the inductionhypothesis to get the desired result?
$endgroup$
– New2Math
19 hours ago
$begingroup$
I think it should be possible to get that to work although it's probably hiding what's going on, rather than illuminating the mechanism at work. However, your comment got me thinking and I've posted an answer below that I think is more straight forward to understand. Hope you like it !
$endgroup$
– Martin Hansen
18 hours ago
$begingroup$
I think it should be possible to get that to work although it's probably hiding what's going on, rather than illuminating the mechanism at work. However, your comment got me thinking and I've posted an answer below that I think is more straight forward to understand. Hope you like it !
$endgroup$
– Martin Hansen
18 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I've been playing around with trying to find a more visual way of proving the Cauchy Product, which amongst other things, multiplies two polynomials together as you've asked for;
My starting point was thinking about a specific example;
$$(2+7x+3x^2)(6+4x+5x^2) =$$
beginarrayc
&2 &7x &3x^2 \ hline
6 &12&42x&18x^2\ hline
4x &8x&28x^2&12x^3\ hline
5x^2 &10x^2&35x^3&15x^4\ hline
endarray
So
$$(2+7x+3x^2)(6+4x+5x^2) = 12+50x+56x^2+47x^3+15x^4$$
Generalising this,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)=$$
beginarrayc
&a_0 &a_1x &a_2x^2 &a_3x^3&dots&a_nx^n \ hline
b_0 &a_0b_0&a_1b_0x&a_2b_0x^2&a_3b_0x^3&dots&a_nb_0x^n\ hline
b_1x &a_0b_1x&a_1b_1x^2&a_2b_1x^3&a_3b_1x^4&dots&a_nb_1x^n+1\ hline
b_2x^2 &a_0b_2x^2&a_1b_2x^3&a_2b_2x^4&a_3b_2x^5&dots&a_nb_2x^n+2\ hline
b_3x^3&a_0b_3x^3&a_1b_3x^4&a_2b_3x^5&a_3b_3x^6&dots&a_nb_3x^n+3\ hline
dots&dots&dots&dots&dots&dots&dots\ hline
b_mx^m&a_0b_mx^m&a_1b_mx^m+1&a_2b_mx^m+2&a_3b_mx^m+3&dots&a_nb_mx^n+m\ hline
endarray
So,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)$$
$$=a_0b_0$$
$$+(a_0b_1+a_1b_0)x$$
$$+(a_0b_2+a_1b_1+a_2b_0)x^2$$
$$+(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3$$
$$+dots$$
$$+a_nb_nx^n+m$$
which is what you were trying to prove.
answered 18 hours ago
Martin HansenMartin Hansen
19113
$endgroup$
$begingroup$
I kinda see the mechanism looked at it for some time now. We generalise it by Looking at a square i.e if $m<n$ than if $n>k>m$ we set $b_k=0$. Then we can prove by induction over $n$ that Picking the $k$th. diagonal from the set of the first $n$ diagonals. We will get $k$ entries. The same goes for the set of the last $n$ diagonals. So we can describe the square by substracting the one double diagonal in the middle which we have in excess. So we have proved the Formula for the square. I am still thinking how one can prove the Formula for a General guadrangle if you know what I mean.
$endgroup$
– New2Math
17 hours ago
add a comment |
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1 Answer
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1 Answer
1
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oldest
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oldest
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oldest
votes
$begingroup$
I've been playing around with trying to find a more visual way of proving the Cauchy Product, which amongst other things, multiplies two polynomials together as you've asked for;
My starting point was thinking about a specific example;
$$(2+7x+3x^2)(6+4x+5x^2) =$$
beginarrayc
&2 &7x &3x^2 \ hline
6 &12&42x&18x^2\ hline
4x &8x&28x^2&12x^3\ hline
5x^2 &10x^2&35x^3&15x^4\ hline
endarray
So
$$(2+7x+3x^2)(6+4x+5x^2) = 12+50x+56x^2+47x^3+15x^4$$
Generalising this,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)=$$
beginarrayc
&a_0 &a_1x &a_2x^2 &a_3x^3&dots&a_nx^n \ hline
b_0 &a_0b_0&a_1b_0x&a_2b_0x^2&a_3b_0x^3&dots&a_nb_0x^n\ hline
b_1x &a_0b_1x&a_1b_1x^2&a_2b_1x^3&a_3b_1x^4&dots&a_nb_1x^n+1\ hline
b_2x^2 &a_0b_2x^2&a_1b_2x^3&a_2b_2x^4&a_3b_2x^5&dots&a_nb_2x^n+2\ hline
b_3x^3&a_0b_3x^3&a_1b_3x^4&a_2b_3x^5&a_3b_3x^6&dots&a_nb_3x^n+3\ hline
dots&dots&dots&dots&dots&dots&dots\ hline
b_mx^m&a_0b_mx^m&a_1b_mx^m+1&a_2b_mx^m+2&a_3b_mx^m+3&dots&a_nb_mx^n+m\ hline
endarray
So,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)$$
$$=a_0b_0$$
$$+(a_0b_1+a_1b_0)x$$
$$+(a_0b_2+a_1b_1+a_2b_0)x^2$$
$$+(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3$$
$$+dots$$
$$+a_nb_nx^n+m$$
which is what you were trying to prove.
answered 18 hours ago
Martin HansenMartin Hansen
19113
$endgroup$
$begingroup$
I kinda see the mechanism looked at it for some time now. We generalise it by Looking at a square i.e if $m<n$ than if $n>k>m$ we set $b_k=0$. Then we can prove by induction over $n$ that Picking the $k$th. diagonal from the set of the first $n$ diagonals. We will get $k$ entries. The same goes for the set of the last $n$ diagonals. So we can describe the square by substracting the one double diagonal in the middle which we have in excess. So we have proved the Formula for the square. I am still thinking how one can prove the Formula for a General guadrangle if you know what I mean.
$endgroup$
– New2Math
17 hours ago
add a comment |
$begingroup$
I've been playing around with trying to find a more visual way of proving the Cauchy Product, which amongst other things, multiplies two polynomials together as you've asked for;
My starting point was thinking about a specific example;
$$(2+7x+3x^2)(6+4x+5x^2) =$$
beginarrayc
&2 &7x &3x^2 \ hline
6 &12&42x&18x^2\ hline
4x &8x&28x^2&12x^3\ hline
5x^2 &10x^2&35x^3&15x^4\ hline
endarray
So
$$(2+7x+3x^2)(6+4x+5x^2) = 12+50x+56x^2+47x^3+15x^4$$
Generalising this,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)=$$
beginarrayc
&a_0 &a_1x &a_2x^2 &a_3x^3&dots&a_nx^n \ hline
b_0 &a_0b_0&a_1b_0x&a_2b_0x^2&a_3b_0x^3&dots&a_nb_0x^n\ hline
b_1x &a_0b_1x&a_1b_1x^2&a_2b_1x^3&a_3b_1x^4&dots&a_nb_1x^n+1\ hline
b_2x^2 &a_0b_2x^2&a_1b_2x^3&a_2b_2x^4&a_3b_2x^5&dots&a_nb_2x^n+2\ hline
b_3x^3&a_0b_3x^3&a_1b_3x^4&a_2b_3x^5&a_3b_3x^6&dots&a_nb_3x^n+3\ hline
dots&dots&dots&dots&dots&dots&dots\ hline
b_mx^m&a_0b_mx^m&a_1b_mx^m+1&a_2b_mx^m+2&a_3b_mx^m+3&dots&a_nb_mx^n+m\ hline
endarray
So,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)$$
$$=a_0b_0$$
$$+(a_0b_1+a_1b_0)x$$
$$+(a_0b_2+a_1b_1+a_2b_0)x^2$$
$$+(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3$$
$$+dots$$
$$+a_nb_nx^n+m$$
which is what you were trying to prove.
answered 18 hours ago
Martin HansenMartin Hansen
19113
$endgroup$
$begingroup$
I kinda see the mechanism looked at it for some time now. We generalise it by Looking at a square i.e if $m<n$ than if $n>k>m$ we set $b_k=0$. Then we can prove by induction over $n$ that Picking the $k$th. diagonal from the set of the first $n$ diagonals. We will get $k$ entries. The same goes for the set of the last $n$ diagonals. So we can describe the square by substracting the one double diagonal in the middle which we have in excess. So we have proved the Formula for the square. I am still thinking how one can prove the Formula for a General guadrangle if you know what I mean.
$endgroup$
– New2Math
17 hours ago
add a comment |
$begingroup$
I've been playing around with trying to find a more visual way of proving the Cauchy Product, which amongst other things, multiplies two polynomials together as you've asked for;
My starting point was thinking about a specific example;
$$(2+7x+3x^2)(6+4x+5x^2) =$$
beginarrayc
&2 &7x &3x^2 \ hline
6 &12&42x&18x^2\ hline
4x &8x&28x^2&12x^3\ hline
5x^2 &10x^2&35x^3&15x^4\ hline
endarray
So
$$(2+7x+3x^2)(6+4x+5x^2) = 12+50x+56x^2+47x^3+15x^4$$
Generalising this,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)=$$
beginarrayc
&a_0 &a_1x &a_2x^2 &a_3x^3&dots&a_nx^n \ hline
b_0 &a_0b_0&a_1b_0x&a_2b_0x^2&a_3b_0x^3&dots&a_nb_0x^n\ hline
b_1x &a_0b_1x&a_1b_1x^2&a_2b_1x^3&a_3b_1x^4&dots&a_nb_1x^n+1\ hline
b_2x^2 &a_0b_2x^2&a_1b_2x^3&a_2b_2x^4&a_3b_2x^5&dots&a_nb_2x^n+2\ hline
b_3x^3&a_0b_3x^3&a_1b_3x^4&a_2b_3x^5&a_3b_3x^6&dots&a_nb_3x^n+3\ hline
dots&dots&dots&dots&dots&dots&dots\ hline
b_mx^m&a_0b_mx^m&a_1b_mx^m+1&a_2b_mx^m+2&a_3b_mx^m+3&dots&a_nb_mx^n+m\ hline
endarray
So,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)$$
$$=a_0b_0$$
$$+(a_0b_1+a_1b_0)x$$
$$+(a_0b_2+a_1b_1+a_2b_0)x^2$$
$$+(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3$$
$$+dots$$
$$+a_nb_nx^n+m$$
which is what you were trying to prove.
answered 18 hours ago
Martin HansenMartin Hansen
19113
$endgroup$
I've been playing around with trying to find a more visual way of proving the Cauchy Product, which amongst other things, multiplies two polynomials together as you've asked for;
My starting point was thinking about a specific example;
$$(2+7x+3x^2)(6+4x+5x^2) =$$
beginarrayc
&2 &7x &3x^2 \ hline
6 &12&42x&18x^2\ hline
4x &8x&28x^2&12x^3\ hline
5x^2 &10x^2&35x^3&15x^4\ hline
endarray
So
$$(2+7x+3x^2)(6+4x+5x^2) = 12+50x+56x^2+47x^3+15x^4$$
Generalising this,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)=$$
beginarrayc
&a_0 &a_1x &a_2x^2 &a_3x^3&dots&a_nx^n \ hline
b_0 &a_0b_0&a_1b_0x&a_2b_0x^2&a_3b_0x^3&dots&a_nb_0x^n\ hline
b_1x &a_0b_1x&a_1b_1x^2&a_2b_1x^3&a_3b_1x^4&dots&a_nb_1x^n+1\ hline
b_2x^2 &a_0b_2x^2&a_1b_2x^3&a_2b_2x^4&a_3b_2x^5&dots&a_nb_2x^n+2\ hline
b_3x^3&a_0b_3x^3&a_1b_3x^4&a_2b_3x^5&a_3b_3x^6&dots&a_nb_3x^n+3\ hline
dots&dots&dots&dots&dots&dots&dots\ hline
b_mx^m&a_0b_mx^m&a_1b_mx^m+1&a_2b_mx^m+2&a_3b_mx^m+3&dots&a_nb_mx^n+m\ hline
endarray
So,
$$(a_0+a_1x+a_2x^2+a_3x^3+dots+a_nx^n)(b_0+a_1x+b_2x^2+b_3x^3+dots+b_mx^m)$$
$$=a_0b_0$$
$$+(a_0b_1+a_1b_0)x$$
$$+(a_0b_2+a_1b_1+a_2b_0)x^2$$
$$+(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3$$
$$+dots$$
$$+a_nb_nx^n+m$$
which is what you were trying to prove.
answered 18 hours ago
Martin HansenMartin Hansen
19113
answered 18 hours ago
Martin HansenMartin Hansen
19113
answered 18 hours ago
Martin HansenMartin Hansen
19113
answered 18 hours ago
Martin HansenMartin Hansen
19113
19113
$begingroup$
I kinda see the mechanism looked at it for some time now. We generalise it by Looking at a square i.e if $m<n$ than if $n>k>m$ we set $b_k=0$. Then we can prove by induction over $n$ that Picking the $k$th. diagonal from the set of the first $n$ diagonals. We will get $k$ entries. The same goes for the set of the last $n$ diagonals. So we can describe the square by substracting the one double diagonal in the middle which we have in excess. So we have proved the Formula for the square. I am still thinking how one can prove the Formula for a General guadrangle if you know what I mean.
$endgroup$
– New2Math
17 hours ago
add a comment |
$begingroup$
I kinda see the mechanism looked at it for some time now. We generalise it by Looking at a square i.e if $m<n$ than if $n>k>m$ we set $b_k=0$. Then we can prove by induction over $n$ that Picking the $k$th. diagonal from the set of the first $n$ diagonals. We will get $k$ entries. The same goes for the set of the last $n$ diagonals. So we can describe the square by substracting the one double diagonal in the middle which we have in excess. So we have proved the Formula for the square. I am still thinking how one can prove the Formula for a General guadrangle if you know what I mean.
$endgroup$
– New2Math
17 hours ago
$begingroup$
I kinda see the mechanism looked at it for some time now. We generalise it by Looking at a square i.e if $m<n$ than if $n>k>m$ we set $b_k=0$. Then we can prove by induction over $n$ that Picking the $k$th. diagonal from the set of the first $n$ diagonals. We will get $k$ entries. The same goes for the set of the last $n$ diagonals. So we can describe the square by substracting the one double diagonal in the middle which we have in excess. So we have proved the Formula for the square. I am still thinking how one can prove the Formula for a General guadrangle if you know what I mean.
$endgroup$
– New2Math
17 hours ago
$begingroup$
I kinda see the mechanism looked at it for some time now. We generalise it by Looking at a square i.e if $m<n$ than if $n>k>m$ we set $b_k=0$. Then we can prove by induction over $n$ that Picking the $k$th. diagonal from the set of the first $n$ diagonals. We will get $k$ entries. The same goes for the set of the last $n$ diagonals. So we can describe the square by substracting the one double diagonal in the middle which we have in excess. So we have proved the Formula for the square. I am still thinking how one can prove the Formula for a General guadrangle if you know what I mean.
$endgroup$
– New2Math
17 hours ago
add a comment |
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$begingroup$
Do you mean fg(x) which is composite functions ? I suspect from the title you mean f(x)g(x). If so, looking up "Cauchy Product" may help.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
@MartinHansen I have tried to prove the Formula with induction First if $m=0,n=0$ the Formula holds. Then Setting $m>0$ and $n=0$ The Formula still holds and then I make an iduction over $n$ and in the inductionstep after using the assocaitive and distributive law I have got the term $a_n+1b_mx^n+1+m+(c_n+1+m-1+a_n+1b_m-1)x^n+1+m-1+...+(c_n+1+a_n+1b_0)x^n+1+c_nx^n+...c_0$. Can you tell me how I have to Phrase the inductionhypothesis to get the desired result?
$endgroup$
– New2Math
19 hours ago
$begingroup$
I think it should be possible to get that to work although it's probably hiding what's going on, rather than illuminating the mechanism at work. However, your comment got me thinking and I've posted an answer below that I think is more straight forward to understand. Hope you like it !
$endgroup$
– Martin Hansen
18 hours ago