Proving the product rule for polynomials without limits or derivative rulesApostol Calculus Vol.1 Exercise 9 , Chapter 1.5 (Prove property of polynomial function)If something holds for n+1 distinct values then it holds for all values.Proving a property of polynomialShow that $(1+a_1x+ldots+a_rx^r)^k=1+x+x^r+1q(x)$Generalization of the derivative to polynomial ringsinduction with remaindersProve that $R(n) =frac2nsum_k=0^n-1 (R(k))+c$ is $R(n) = n*c$Proving by induction that order of the product is equal to the product of the ordersExpressing Laurent polynomials using generators and relationsPolynomial Length upper boundProving the connection between the Binomial Theorem and the product rule for derivatives
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Proving the product rule for polynomials without limits or derivative rules
Apostol Calculus Vol.1 Exercise 9 , Chapter 1.5 (Prove property of polynomial function)If something holds for n+1 distinct values then it holds for all values.Proving a property of polynomialShow that $(1+a_1x+ldots+a_rx^r)^k=1+x+x^r+1q(x)$Generalization of the derivative to polynomial ringsinduction with remaindersProve that $R(n) =frac2nsum_k=0^n-1 (R(k))+c$ is $R(n) = n*c$Proving by induction that order of the product is equal to the product of the ordersExpressing Laurent polynomials using generators and relationsPolynomial Length upper boundProving the connection between the Binomial Theorem and the product rule for derivatives
$begingroup$
As I said in the title, in the book where this exercise was proposed, derivatives or limits were not introduced yet. The exercise was to show that given the function $':mathbbC[x]rightarrow mathbbC[x]$ I should show that $(fg)'=f'g+fg'$
$f'(x)$ was defined as $sum_k=1^nka_kx^k-1$ where $f(x)=sum_k=0^na_kx^k$.
I have found out that an equivalent representation for $f'(x)$ is $sum_k=0^n-1(k+1)a_k+1x^k$
I have tried to prove the result myself step by step but there must be an error somewhere in the proof because in the end the result does not fit the bill.
$fg=sum_k=0^m+nc_kx^k$ with $c_k=sum_r+s=ka_rb_s$
$(fg)'=sum_j=0^m+n-1(j+1)c_j+1x^j$
$f=sum_r=0^na_rx^r$
$f'=sum_r=0^n-1(r+1)a_r+1x^r$
$g=sum_s=0^mb_sx^s$
$g'=sum_s=0^m-1(s+1)b_s+1x^s$
$f'g=sum_j=0^m+n-1c'_jx^j$, with $c'_j=sum_r+s=j(r+1)a_r+1b_s$
$fg'=sum_j=0^m+n-1c''_jx^j$, with $c''_j=sum_r+s=ja_r(s+1)b_s+1$
I have tried to prove by induction that for every $j$ we have
$c'_j+c''_j=(j+1)c_j+1$
but I failed to show that the coefficients of both polynomials are equal. Where is the mistake? Can somebody give me a hint please?
The Problem is the induction step; I don't know how I can use the induction hypothesis to my advantage; I have made the induction over $j$ and I could Show it for $j=0$ and $j=1$ (just to be sure):
The claim is
$$(j+1)sum_r+s=j+1a_rb_s=sum_r+s=j(r+1)a_r+1b_s+sum_r+s=ja_r(s+1)b_s+1$$
I also know that $(j+1)sum_r+s=j+1a_rb_s=(j+1)sum_r=0^j+1a_rb_j+1-r$
polynomials complex-numbers induction
edited Mar 12 at 21:09
J. W. Tanner
3,3351320
asked Mar 12 at 20:12
New2MathNew2Math
12312
$endgroup$
add a comment |
$begingroup$
As I said in the title, in the book where this exercise was proposed, derivatives or limits were not introduced yet. The exercise was to show that given the function $':mathbbC[x]rightarrow mathbbC[x]$ I should show that $(fg)'=f'g+fg'$
$f'(x)$ was defined as $sum_k=1^nka_kx^k-1$ where $f(x)=sum_k=0^na_kx^k$.
I have found out that an equivalent representation for $f'(x)$ is $sum_k=0^n-1(k+1)a_k+1x^k$
I have tried to prove the result myself step by step but there must be an error somewhere in the proof because in the end the result does not fit the bill.
$fg=sum_k=0^m+nc_kx^k$ with $c_k=sum_r+s=ka_rb_s$
$(fg)'=sum_j=0^m+n-1(j+1)c_j+1x^j$
$f=sum_r=0^na_rx^r$
$f'=sum_r=0^n-1(r+1)a_r+1x^r$
$g=sum_s=0^mb_sx^s$
$g'=sum_s=0^m-1(s+1)b_s+1x^s$
$f'g=sum_j=0^m+n-1c'_jx^j$, with $c'_j=sum_r+s=j(r+1)a_r+1b_s$
$fg'=sum_j=0^m+n-1c''_jx^j$, with $c''_j=sum_r+s=ja_r(s+1)b_s+1$
I have tried to prove by induction that for every $j$ we have
$c'_j+c''_j=(j+1)c_j+1$
but I failed to show that the coefficients of both polynomials are equal. Where is the mistake? Can somebody give me a hint please?
The Problem is the induction step; I don't know how I can use the induction hypothesis to my advantage; I have made the induction over $j$ and I could Show it for $j=0$ and $j=1$ (just to be sure):
The claim is
$$(j+1)sum_r+s=j+1a_rb_s=sum_r+s=j(r+1)a_r+1b_s+sum_r+s=ja_r(s+1)b_s+1$$
I also know that $(j+1)sum_r+s=j+1a_rb_s=(j+1)sum_r=0^j+1a_rb_j+1-r$
polynomials complex-numbers induction
edited Mar 12 at 21:09
J. W. Tanner
3,3351320
asked Mar 12 at 20:12
New2MathNew2Math
12312
$endgroup$
$begingroup$
Try proving it first for monomials (the case $f(x) = x^r$, $g(x) = x^s$) and then use linearity to prove the general case.
$endgroup$
– Jair Taylor
Mar 12 at 21:08
add a comment |
$begingroup$
As I said in the title, in the book where this exercise was proposed, derivatives or limits were not introduced yet. The exercise was to show that given the function $':mathbbC[x]rightarrow mathbbC[x]$ I should show that $(fg)'=f'g+fg'$
$f'(x)$ was defined as $sum_k=1^nka_kx^k-1$ where $f(x)=sum_k=0^na_kx^k$.
I have found out that an equivalent representation for $f'(x)$ is $sum_k=0^n-1(k+1)a_k+1x^k$
I have tried to prove the result myself step by step but there must be an error somewhere in the proof because in the end the result does not fit the bill.
$fg=sum_k=0^m+nc_kx^k$ with $c_k=sum_r+s=ka_rb_s$
$(fg)'=sum_j=0^m+n-1(j+1)c_j+1x^j$
$f=sum_r=0^na_rx^r$
$f'=sum_r=0^n-1(r+1)a_r+1x^r$
$g=sum_s=0^mb_sx^s$
$g'=sum_s=0^m-1(s+1)b_s+1x^s$
$f'g=sum_j=0^m+n-1c'_jx^j$, with $c'_j=sum_r+s=j(r+1)a_r+1b_s$
$fg'=sum_j=0^m+n-1c''_jx^j$, with $c''_j=sum_r+s=ja_r(s+1)b_s+1$
I have tried to prove by induction that for every $j$ we have
$c'_j+c''_j=(j+1)c_j+1$
but I failed to show that the coefficients of both polynomials are equal. Where is the mistake? Can somebody give me a hint please?
The Problem is the induction step; I don't know how I can use the induction hypothesis to my advantage; I have made the induction over $j$ and I could Show it for $j=0$ and $j=1$ (just to be sure):
The claim is
$$(j+1)sum_r+s=j+1a_rb_s=sum_r+s=j(r+1)a_r+1b_s+sum_r+s=ja_r(s+1)b_s+1$$
I also know that $(j+1)sum_r+s=j+1a_rb_s=(j+1)sum_r=0^j+1a_rb_j+1-r$
polynomials complex-numbers induction
edited Mar 12 at 21:09
J. W. Tanner
3,3351320
asked Mar 12 at 20:12
New2MathNew2Math
12312
$endgroup$
As I said in the title, in the book where this exercise was proposed, derivatives or limits were not introduced yet. The exercise was to show that given the function $':mathbbC[x]rightarrow mathbbC[x]$ I should show that $(fg)'=f'g+fg'$
$f'(x)$ was defined as $sum_k=1^nka_kx^k-1$ where $f(x)=sum_k=0^na_kx^k$.
I have found out that an equivalent representation for $f'(x)$ is $sum_k=0^n-1(k+1)a_k+1x^k$
I have tried to prove the result myself step by step but there must be an error somewhere in the proof because in the end the result does not fit the bill.
$fg=sum_k=0^m+nc_kx^k$ with $c_k=sum_r+s=ka_rb_s$
$(fg)'=sum_j=0^m+n-1(j+1)c_j+1x^j$
$f=sum_r=0^na_rx^r$
$f'=sum_r=0^n-1(r+1)a_r+1x^r$
$g=sum_s=0^mb_sx^s$
$g'=sum_s=0^m-1(s+1)b_s+1x^s$
$f'g=sum_j=0^m+n-1c'_jx^j$, with $c'_j=sum_r+s=j(r+1)a_r+1b_s$
$fg'=sum_j=0^m+n-1c''_jx^j$, with $c''_j=sum_r+s=ja_r(s+1)b_s+1$
I have tried to prove by induction that for every $j$ we have
$c'_j+c''_j=(j+1)c_j+1$
but I failed to show that the coefficients of both polynomials are equal. Where is the mistake? Can somebody give me a hint please?
The Problem is the induction step; I don't know how I can use the induction hypothesis to my advantage; I have made the induction over $j$ and I could Show it for $j=0$ and $j=1$ (just to be sure):
The claim is
$$(j+1)sum_r+s=j+1a_rb_s=sum_r+s=j(r+1)a_r+1b_s+sum_r+s=ja_r(s+1)b_s+1$$
I also know that $(j+1)sum_r+s=j+1a_rb_s=(j+1)sum_r=0^j+1a_rb_j+1-r$
polynomials complex-numbers induction
polynomials complex-numbers induction
edited Mar 12 at 21:09
J. W. Tanner
3,3351320
asked Mar 12 at 20:12
New2MathNew2Math
12312
edited Mar 12 at 21:09
J. W. Tanner
3,3351320
asked Mar 12 at 20:12
New2MathNew2Math
12312
edited Mar 12 at 21:09
J. W. Tanner
3,3351320
edited Mar 12 at 21:09
J. W. Tanner
3,3351320
edited Mar 12 at 21:09
J. W. Tanner
3,3351320
3,3351320
asked Mar 12 at 20:12
New2MathNew2Math
12312
asked Mar 12 at 20:12
New2MathNew2Math
12312
asked Mar 12 at 20:12
New2MathNew2Math
12312
12312
$begingroup$
Try proving it first for monomials (the case $f(x) = x^r$, $g(x) = x^s$) and then use linearity to prove the general case.
$endgroup$
– Jair Taylor
Mar 12 at 21:08
add a comment |
$begingroup$
Try proving it first for monomials (the case $f(x) = x^r$, $g(x) = x^s$) and then use linearity to prove the general case.
$endgroup$
– Jair Taylor
Mar 12 at 21:08
$begingroup$
Try proving it first for monomials (the case $f(x) = x^r$, $g(x) = x^s$) and then use linearity to prove the general case.
$endgroup$
– Jair Taylor
Mar 12 at 21:08
$begingroup$
Try proving it first for monomials (the case $f(x) = x^r$, $g(x) = x^s$) and then use linearity to prove the general case.
$endgroup$
– Jair Taylor
Mar 12 at 21:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is easier to not use coefficients at all - except that we initially check by straightforward computation
$$tag1 (f+g)'=f'+g',$$ $$tag2 (cf)'=cf',$$ $$tag3 (xf)'=f+f'.$$
$$tag 4c'=0$$
Let $Asubseteq Bbb C[x]$ be the set of polynomials $f$ with the property that $(fg)'=f'g+fg'$ for all polynomials $g$.
If $f_1,f_2in A$, then for every polynomial $g$,
$$beginalign((f_1+f_2)g)'&=(f_1g+f_2g)'\&=(f_1g)'+(f_2g)'&textby (1)\&=f_1g'+f_1'g+f_2g'+f_2'g\&=(f_1+f_2)g'+(f_1'+f_2')g\
&=(f_1+f_2)g'+(f_1+f_2)'g&textby (1)endalign $$
Thus,
$A$ is closed under addition.
Assume $f_1,f_2in A$. Then for any polynomial $g$,
$$beginalign
((f_1f_2)g)'&=(f_1(f_2g))'\
&=f_1'f_2g+f_1(f_2g)'\
&=f_1'f_2g+f_1(f_2'g+f_2g')\
&=(f_1'f_2+f_1f_2')g+f_1f_2g'.
endalign$$
We conclude that
$A$ is closed under multiplication.
From $(2)$ and $(4)$, we see that
$A$ contains all constants
and from $(3)$ (after verifying that in particular $x'=1$)
$xin A$.
Now observe that the four highlighted statements imply that $A=Bbb C[x]$.
answered Mar 12 at 22:02
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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votes
$begingroup$
It is easier to not use coefficients at all - except that we initially check by straightforward computation
$$tag1 (f+g)'=f'+g',$$ $$tag2 (cf)'=cf',$$ $$tag3 (xf)'=f+f'.$$
$$tag 4c'=0$$
Let $Asubseteq Bbb C[x]$ be the set of polynomials $f$ with the property that $(fg)'=f'g+fg'$ for all polynomials $g$.
If $f_1,f_2in A$, then for every polynomial $g$,
$$beginalign((f_1+f_2)g)'&=(f_1g+f_2g)'\&=(f_1g)'+(f_2g)'&textby (1)\&=f_1g'+f_1'g+f_2g'+f_2'g\&=(f_1+f_2)g'+(f_1'+f_2')g\
&=(f_1+f_2)g'+(f_1+f_2)'g&textby (1)endalign $$
Thus,
$A$ is closed under addition.
Assume $f_1,f_2in A$. Then for any polynomial $g$,
$$beginalign
((f_1f_2)g)'&=(f_1(f_2g))'\
&=f_1'f_2g+f_1(f_2g)'\
&=f_1'f_2g+f_1(f_2'g+f_2g')\
&=(f_1'f_2+f_1f_2')g+f_1f_2g'.
endalign$$
We conclude that
$A$ is closed under multiplication.
From $(2)$ and $(4)$, we see that
$A$ contains all constants
and from $(3)$ (after verifying that in particular $x'=1$)
$xin A$.
Now observe that the four highlighted statements imply that $A=Bbb C[x]$.
answered Mar 12 at 22:02
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
It is easier to not use coefficients at all - except that we initially check by straightforward computation
$$tag1 (f+g)'=f'+g',$$ $$tag2 (cf)'=cf',$$ $$tag3 (xf)'=f+f'.$$
$$tag 4c'=0$$
Let $Asubseteq Bbb C[x]$ be the set of polynomials $f$ with the property that $(fg)'=f'g+fg'$ for all polynomials $g$.
If $f_1,f_2in A$, then for every polynomial $g$,
$$beginalign((f_1+f_2)g)'&=(f_1g+f_2g)'\&=(f_1g)'+(f_2g)'&textby (1)\&=f_1g'+f_1'g+f_2g'+f_2'g\&=(f_1+f_2)g'+(f_1'+f_2')g\
&=(f_1+f_2)g'+(f_1+f_2)'g&textby (1)endalign $$
Thus,
$A$ is closed under addition.
Assume $f_1,f_2in A$. Then for any polynomial $g$,
$$beginalign
((f_1f_2)g)'&=(f_1(f_2g))'\
&=f_1'f_2g+f_1(f_2g)'\
&=f_1'f_2g+f_1(f_2'g+f_2g')\
&=(f_1'f_2+f_1f_2')g+f_1f_2g'.
endalign$$
We conclude that
$A$ is closed under multiplication.
From $(2)$ and $(4)$, we see that
$A$ contains all constants
and from $(3)$ (after verifying that in particular $x'=1$)
$xin A$.
Now observe that the four highlighted statements imply that $A=Bbb C[x]$.
answered Mar 12 at 22:02
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
It is easier to not use coefficients at all - except that we initially check by straightforward computation
$$tag1 (f+g)'=f'+g',$$ $$tag2 (cf)'=cf',$$ $$tag3 (xf)'=f+f'.$$
$$tag 4c'=0$$
Let $Asubseteq Bbb C[x]$ be the set of polynomials $f$ with the property that $(fg)'=f'g+fg'$ for all polynomials $g$.
If $f_1,f_2in A$, then for every polynomial $g$,
$$beginalign((f_1+f_2)g)'&=(f_1g+f_2g)'\&=(f_1g)'+(f_2g)'&textby (1)\&=f_1g'+f_1'g+f_2g'+f_2'g\&=(f_1+f_2)g'+(f_1'+f_2')g\
&=(f_1+f_2)g'+(f_1+f_2)'g&textby (1)endalign $$
Thus,
$A$ is closed under addition.
Assume $f_1,f_2in A$. Then for any polynomial $g$,
$$beginalign
((f_1f_2)g)'&=(f_1(f_2g))'\
&=f_1'f_2g+f_1(f_2g)'\
&=f_1'f_2g+f_1(f_2'g+f_2g')\
&=(f_1'f_2+f_1f_2')g+f_1f_2g'.
endalign$$
We conclude that
$A$ is closed under multiplication.
From $(2)$ and $(4)$, we see that
$A$ contains all constants
and from $(3)$ (after verifying that in particular $x'=1$)
$xin A$.
Now observe that the four highlighted statements imply that $A=Bbb C[x]$.
answered Mar 12 at 22:02
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
It is easier to not use coefficients at all - except that we initially check by straightforward computation
$$tag1 (f+g)'=f'+g',$$ $$tag2 (cf)'=cf',$$ $$tag3 (xf)'=f+f'.$$
$$tag 4c'=0$$
Let $Asubseteq Bbb C[x]$ be the set of polynomials $f$ with the property that $(fg)'=f'g+fg'$ for all polynomials $g$.
If $f_1,f_2in A$, then for every polynomial $g$,
$$beginalign((f_1+f_2)g)'&=(f_1g+f_2g)'\&=(f_1g)'+(f_2g)'&textby (1)\&=f_1g'+f_1'g+f_2g'+f_2'g\&=(f_1+f_2)g'+(f_1'+f_2')g\
&=(f_1+f_2)g'+(f_1+f_2)'g&textby (1)endalign $$
Thus,
$A$ is closed under addition.
Assume $f_1,f_2in A$. Then for any polynomial $g$,
$$beginalign
((f_1f_2)g)'&=(f_1(f_2g))'\
&=f_1'f_2g+f_1(f_2g)'\
&=f_1'f_2g+f_1(f_2'g+f_2g')\
&=(f_1'f_2+f_1f_2')g+f_1f_2g'.
endalign$$
We conclude that
$A$ is closed under multiplication.
From $(2)$ and $(4)$, we see that
$A$ contains all constants
and from $(3)$ (after verifying that in particular $x'=1$)
$xin A$.
Now observe that the four highlighted statements imply that $A=Bbb C[x]$.
answered Mar 12 at 22:02
Hagen von EitzenHagen von Eitzen
283k23272507
answered Mar 12 at 22:02
Hagen von EitzenHagen von Eitzen
283k23272507
answered Mar 12 at 22:02
Hagen von EitzenHagen von Eitzen
283k23272507
answered Mar 12 at 22:02
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
add a comment |
add a comment |
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$begingroup$
Try proving it first for monomials (the case $f(x) = x^r$, $g(x) = x^s$) and then use linearity to prove the general case.
$endgroup$
– Jair Taylor
Mar 12 at 21:08