Show that if $P(x)$ is an polynomial with an uneven degree then there exists a $x_0in mathbb{R}: P(x_0) <...
$begingroup$
I have eddited the Question, I hope the Statement is now true.
I want to show that every such polynomial has a zeropoint, If I can prove the Statement above, I can use the indermediate value Theorem.
What I have worked out so far is that
$P(x)$, can be rewritten in the form of:
$P(x)= a_0x^{2k+1}+a_1x^{i_1}+a_2x^{i_2}+...a_jx^{i_j}+C,jinmathbb{N}_0$ and $2k+1>i_1>…>i_j$
$= (a_0x^{2k+1-i_j}+a_1x^{i_1-i_j}+...+a_j)x^{i_j}+C$
$= ((a_0x^{2k+1-i_j-(i_{j-1}-i_j)}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$=((a_0x^{2k+1-i_{j-1}}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$...$
$=(…(a_0x^{2k+1-i_1}+a_1)x^{i_2-i_1}+a_2)…)x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
How can I continue, do you have any suggestions?
I had the idea to pick
$maxlimits_{0 leq s leq j}|a_s|$ as my $x$
I somehow think that $P(x)$ would either be positive or negative dependant whether I pick $+x$ or $-x$. If that goes in the right direction I need somebody to tell me how to proceed.
real-analysis polynomials
asked Jan 11 at 16:00
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
I have eddited the Question, I hope the Statement is now true.
I want to show that every such polynomial has a zeropoint, If I can prove the Statement above, I can use the indermediate value Theorem.
What I have worked out so far is that
$P(x)$, can be rewritten in the form of:
$P(x)= a_0x^{2k+1}+a_1x^{i_1}+a_2x^{i_2}+...a_jx^{i_j}+C,jinmathbb{N}_0$ and $2k+1>i_1>…>i_j$
$= (a_0x^{2k+1-i_j}+a_1x^{i_1-i_j}+...+a_j)x^{i_j}+C$
$= ((a_0x^{2k+1-i_j-(i_{j-1}-i_j)}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$=((a_0x^{2k+1-i_{j-1}}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$...$
$=(…(a_0x^{2k+1-i_1}+a_1)x^{i_2-i_1}+a_2)…)x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
How can I continue, do you have any suggestions?
I had the idea to pick
$maxlimits_{0 leq s leq j}|a_s|$ as my $x$
I somehow think that $P(x)$ would either be positive or negative dependant whether I pick $+x$ or $-x$. If that goes in the right direction I need somebody to tell me how to proceed.
real-analysis polynomials
asked Jan 11 at 16:00
RM777RM777
38312
$endgroup$
$begingroup$
You want to use this to prove the intermediate value theorem?
$endgroup$
– Mindlack
Jan 11 at 16:05
$begingroup$
Your statement is imprecise. The polynomial $P(x)=x$ has odd degree $1$ but does not have a positive or a negative zero.
$endgroup$
– MPW
Jan 11 at 16:05
1
$begingroup$
The claim in your title is not true. What are you actually trying to prove?
$endgroup$
– user3482749
Jan 11 at 16:10
$begingroup$
No what I want to show is that every polynomial of uneven Degree has a zeropoint. If I can show that for every such polynomial there exists a positive and a negative value i.e. $exists_{z,z'inmathbb{R}}:P(z)>0,P(z')<0$ I can make use of the intemediate value Theorem to say that $exists x$ in the intervall $[z,z']$ or $([z',z])$: $P(x)=0$. Therefor I assume without loss of generality that $P(x)>0,forall xin mathbb{R}$ and try to invoke a contradiction by constructing a $x$ with negative value.
$endgroup$
– RM777
Jan 11 at 16:11
$begingroup$
look at the limits. (if $P in mathbb{R}[X]$) it should give you the answer
$endgroup$
– Alexis
Jan 11 at 16:15
add a comment |
$begingroup$
I have eddited the Question, I hope the Statement is now true.
I want to show that every such polynomial has a zeropoint, If I can prove the Statement above, I can use the indermediate value Theorem.
What I have worked out so far is that
$P(x)$, can be rewritten in the form of:
$P(x)= a_0x^{2k+1}+a_1x^{i_1}+a_2x^{i_2}+...a_jx^{i_j}+C,jinmathbb{N}_0$ and $2k+1>i_1>…>i_j$
$= (a_0x^{2k+1-i_j}+a_1x^{i_1-i_j}+...+a_j)x^{i_j}+C$
$= ((a_0x^{2k+1-i_j-(i_{j-1}-i_j)}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$=((a_0x^{2k+1-i_{j-1}}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$...$
$=(…(a_0x^{2k+1-i_1}+a_1)x^{i_2-i_1}+a_2)…)x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
How can I continue, do you have any suggestions?
I had the idea to pick
$maxlimits_{0 leq s leq j}|a_s|$ as my $x$
I somehow think that $P(x)$ would either be positive or negative dependant whether I pick $+x$ or $-x$. If that goes in the right direction I need somebody to tell me how to proceed.
real-analysis polynomials
asked Jan 11 at 16:00
RM777RM777
38312
$endgroup$
I have eddited the Question, I hope the Statement is now true.
I want to show that every such polynomial has a zeropoint, If I can prove the Statement above, I can use the indermediate value Theorem.
What I have worked out so far is that
$P(x)$, can be rewritten in the form of:
$P(x)= a_0x^{2k+1}+a_1x^{i_1}+a_2x^{i_2}+...a_jx^{i_j}+C,jinmathbb{N}_0$ and $2k+1>i_1>…>i_j$
$= (a_0x^{2k+1-i_j}+a_1x^{i_1-i_j}+...+a_j)x^{i_j}+C$
$= ((a_0x^{2k+1-i_j-(i_{j-1}-i_j)}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$=((a_0x^{2k+1-i_{j-1}}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
$...$
$=(…(a_0x^{2k+1-i_1}+a_1)x^{i_2-i_1}+a_2)…)x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$
How can I continue, do you have any suggestions?
I had the idea to pick
$maxlimits_{0 leq s leq j}|a_s|$ as my $x$
I somehow think that $P(x)$ would either be positive or negative dependant whether I pick $+x$ or $-x$. If that goes in the right direction I need somebody to tell me how to proceed.
real-analysis polynomials
real-analysis polynomials
asked Jan 11 at 16:00
RM777RM777
38312
asked Jan 11 at 16:00
RM777RM777
38312
edited Jan 11 at 16:19
RM777
asked Jan 11 at 16:00
RM777RM777
38312
asked Jan 11 at 16:00
RM777RM777
38312
asked Jan 11 at 16:00
RM777RM777
38312
38312
$begingroup$
You want to use this to prove the intermediate value theorem?
$endgroup$
– Mindlack
Jan 11 at 16:05
$begingroup$
Your statement is imprecise. The polynomial $P(x)=x$ has odd degree $1$ but does not have a positive or a negative zero.
$endgroup$
– MPW
Jan 11 at 16:05
1
$begingroup$
The claim in your title is not true. What are you actually trying to prove?
$endgroup$
– user3482749
Jan 11 at 16:10
$begingroup$
No what I want to show is that every polynomial of uneven Degree has a zeropoint. If I can show that for every such polynomial there exists a positive and a negative value i.e. $exists_{z,z'inmathbb{R}}:P(z)>0,P(z')<0$ I can make use of the intemediate value Theorem to say that $exists x$ in the intervall $[z,z']$ or $([z',z])$: $P(x)=0$. Therefor I assume without loss of generality that $P(x)>0,forall xin mathbb{R}$ and try to invoke a contradiction by constructing a $x$ with negative value.
$endgroup$
– RM777
Jan 11 at 16:11
$begingroup$
look at the limits. (if $P in mathbb{R}[X]$) it should give you the answer
$endgroup$
– Alexis
Jan 11 at 16:15
add a comment |
$begingroup$
You want to use this to prove the intermediate value theorem?
$endgroup$
– Mindlack
Jan 11 at 16:05
$begingroup$
Your statement is imprecise. The polynomial $P(x)=x$ has odd degree $1$ but does not have a positive or a negative zero.
$endgroup$
– MPW
Jan 11 at 16:05
1
$begingroup$
The claim in your title is not true. What are you actually trying to prove?
$endgroup$
– user3482749
Jan 11 at 16:10
$begingroup$
No what I want to show is that every polynomial of uneven Degree has a zeropoint. If I can show that for every such polynomial there exists a positive and a negative value i.e. $exists_{z,z'inmathbb{R}}:P(z)>0,P(z')<0$ I can make use of the intemediate value Theorem to say that $exists x$ in the intervall $[z,z']$ or $([z',z])$: $P(x)=0$. Therefor I assume without loss of generality that $P(x)>0,forall xin mathbb{R}$ and try to invoke a contradiction by constructing a $x$ with negative value.
$endgroup$
– RM777
Jan 11 at 16:11
$begingroup$
look at the limits. (if $P in mathbb{R}[X]$) it should give you the answer
$endgroup$
– Alexis
Jan 11 at 16:15
$begingroup$
You want to use this to prove the intermediate value theorem?
$endgroup$
– Mindlack
Jan 11 at 16:05
$begingroup$
You want to use this to prove the intermediate value theorem?
$endgroup$
– Mindlack
Jan 11 at 16:05
$begingroup$
Your statement is imprecise. The polynomial $P(x)=x$ has odd degree $1$ but does not have a positive or a negative zero.
$endgroup$
– MPW
Jan 11 at 16:05
$begingroup$
Your statement is imprecise. The polynomial $P(x)=x$ has odd degree $1$ but does not have a positive or a negative zero.
$endgroup$
– MPW
Jan 11 at 16:05
1
1
$begingroup$
The claim in your title is not true. What are you actually trying to prove?
$endgroup$
– user3482749
Jan 11 at 16:10
$begingroup$
The claim in your title is not true. What are you actually trying to prove?
$endgroup$
– user3482749
Jan 11 at 16:10
$begingroup$
No what I want to show is that every polynomial of uneven Degree has a zeropoint. If I can show that for every such polynomial there exists a positive and a negative value i.e. $exists_{z,z'inmathbb{R}}:P(z)>0,P(z')<0$ I can make use of the intemediate value Theorem to say that $exists x$ in the intervall $[z,z']$ or $([z',z])$: $P(x)=0$. Therefor I assume without loss of generality that $P(x)>0,forall xin mathbb{R}$ and try to invoke a contradiction by constructing a $x$ with negative value.
$endgroup$
– RM777
Jan 11 at 16:11
$begingroup$
No what I want to show is that every polynomial of uneven Degree has a zeropoint. If I can show that for every such polynomial there exists a positive and a negative value i.e. $exists_{z,z'inmathbb{R}}:P(z)>0,P(z')<0$ I can make use of the intemediate value Theorem to say that $exists x$ in the intervall $[z,z']$ or $([z',z])$: $P(x)=0$. Therefor I assume without loss of generality that $P(x)>0,forall xin mathbb{R}$ and try to invoke a contradiction by constructing a $x$ with negative value.
$endgroup$
– RM777
Jan 11 at 16:11
$begingroup$
look at the limits. (if $P in mathbb{R}[X]$) it should give you the answer
$endgroup$
– Alexis
Jan 11 at 16:15
$begingroup$
look at the limits. (if $P in mathbb{R}[X]$) it should give you the answer
$endgroup$
– Alexis
Jan 11 at 16:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assume WLOG that $a_{2n+1}gt0$.
We have $p(x)=a_{2n+1}x^{2n+1}+dots+a_0$. But, as you take the limit as $x$ approaches $pminfty$, the leading term dominates. That is, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.
answered Jan 11 at 16:52
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Why can you Claim that so easily don't you have to give an explicit $a$ such that $forall_{rinmathbb{R}}exists_{ainmathbb{R}}forall_{xinmathbb{R}}:xin(a,+infty)Rightarrow p(x)>r$
$endgroup$
– RM777
Jan 11 at 19:32
$begingroup$
But no worries I could solve and understand the Problem now.
$endgroup$
– RM777
Jan 11 at 19:40
$begingroup$
Well that's fairly easy to do; given that the expression in parentheses approaches $1$. That is, take $x$ such that both $x^{2n+1}gtfrac r{a_{2n+1}}$ and the expression in parentheses is sufficiently close to $1$. There are some details to consider, but it clearly works out.
$endgroup$
– Chris Custer
Jan 11 at 19:45
$begingroup$
Sorry I thought I had it but I still don't understand it why does the expression in the parentheses approach 1? Take for example $frac{a_0}{x^{2n+1}}$ it is true if I take a big $x$ then the term goes as close to 0 as I want but I if I multiply this value by$ x^{2n+1}$ then I have$ a_0 $again which does not have to be $1$
$endgroup$
– RM777
Jan 11 at 20:05
$begingroup$
Yeah, typo on my part. The expression in parentheses approaches $a_{2n+1}$. And you have given the reason why: the other terms all go to zero.
$endgroup$
– Chris Custer
Jan 11 at 20:08
|
show 4 more comments
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Assume WLOG that $a_{2n+1}gt0$.
We have $p(x)=a_{2n+1}x^{2n+1}+dots+a_0$. But, as you take the limit as $x$ approaches $pminfty$, the leading term dominates. That is, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.
answered Jan 11 at 16:52
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Why can you Claim that so easily don't you have to give an explicit $a$ such that $forall_{rinmathbb{R}}exists_{ainmathbb{R}}forall_{xinmathbb{R}}:xin(a,+infty)Rightarrow p(x)>r$
$endgroup$
– RM777
Jan 11 at 19:32
$begingroup$
But no worries I could solve and understand the Problem now.
$endgroup$
– RM777
Jan 11 at 19:40
$begingroup$
Well that's fairly easy to do; given that the expression in parentheses approaches $1$. That is, take $x$ such that both $x^{2n+1}gtfrac r{a_{2n+1}}$ and the expression in parentheses is sufficiently close to $1$. There are some details to consider, but it clearly works out.
$endgroup$
– Chris Custer
Jan 11 at 19:45
$begingroup$
Sorry I thought I had it but I still don't understand it why does the expression in the parentheses approach 1? Take for example $frac{a_0}{x^{2n+1}}$ it is true if I take a big $x$ then the term goes as close to 0 as I want but I if I multiply this value by$ x^{2n+1}$ then I have$ a_0 $again which does not have to be $1$
$endgroup$
– RM777
Jan 11 at 20:05
$begingroup$
Yeah, typo on my part. The expression in parentheses approaches $a_{2n+1}$. And you have given the reason why: the other terms all go to zero.
$endgroup$
– Chris Custer
Jan 11 at 20:08
|
show 4 more comments
$begingroup$
Assume WLOG that $a_{2n+1}gt0$.
We have $p(x)=a_{2n+1}x^{2n+1}+dots+a_0$. But, as you take the limit as $x$ approaches $pminfty$, the leading term dominates. That is, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.
answered Jan 11 at 16:52
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Why can you Claim that so easily don't you have to give an explicit $a$ such that $forall_{rinmathbb{R}}exists_{ainmathbb{R}}forall_{xinmathbb{R}}:xin(a,+infty)Rightarrow p(x)>r$
$endgroup$
– RM777
Jan 11 at 19:32
$begingroup$
But no worries I could solve and understand the Problem now.
$endgroup$
– RM777
Jan 11 at 19:40
$begingroup$
Well that's fairly easy to do; given that the expression in parentheses approaches $1$. That is, take $x$ such that both $x^{2n+1}gtfrac r{a_{2n+1}}$ and the expression in parentheses is sufficiently close to $1$. There are some details to consider, but it clearly works out.
$endgroup$
– Chris Custer
Jan 11 at 19:45
$begingroup$
Sorry I thought I had it but I still don't understand it why does the expression in the parentheses approach 1? Take for example $frac{a_0}{x^{2n+1}}$ it is true if I take a big $x$ then the term goes as close to 0 as I want but I if I multiply this value by$ x^{2n+1}$ then I have$ a_0 $again which does not have to be $1$
$endgroup$
– RM777
Jan 11 at 20:05
$begingroup$
Yeah, typo on my part. The expression in parentheses approaches $a_{2n+1}$. And you have given the reason why: the other terms all go to zero.
$endgroup$
– Chris Custer
Jan 11 at 20:08
|
show 4 more comments
$begingroup$
Assume WLOG that $a_{2n+1}gt0$.
We have $p(x)=a_{2n+1}x^{2n+1}+dots+a_0$. But, as you take the limit as $x$ approaches $pminfty$, the leading term dominates. That is, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.
answered Jan 11 at 16:52
Chris CusterChris Custer
14.2k3827
$endgroup$
Assume WLOG that $a_{2n+1}gt0$.
We have $p(x)=a_{2n+1}x^{2n+1}+dots+a_0$. But, as you take the limit as $x$ approaches $pminfty$, the leading term dominates. That is, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.
answered Jan 11 at 16:52
Chris CusterChris Custer
14.2k3827
answered Jan 11 at 16:52
Chris CusterChris Custer
14.2k3827
answered Jan 11 at 16:52
Chris CusterChris Custer
14.2k3827
answered Jan 11 at 16:52
Chris CusterChris Custer
14.2k3827
14.2k3827
$begingroup$
Why can you Claim that so easily don't you have to give an explicit $a$ such that $forall_{rinmathbb{R}}exists_{ainmathbb{R}}forall_{xinmathbb{R}}:xin(a,+infty)Rightarrow p(x)>r$
$endgroup$
– RM777
Jan 11 at 19:32
$begingroup$
But no worries I could solve and understand the Problem now.
$endgroup$
– RM777
Jan 11 at 19:40
$begingroup$
Well that's fairly easy to do; given that the expression in parentheses approaches $1$. That is, take $x$ such that both $x^{2n+1}gtfrac r{a_{2n+1}}$ and the expression in parentheses is sufficiently close to $1$. There are some details to consider, but it clearly works out.
$endgroup$
– Chris Custer
Jan 11 at 19:45
$begingroup$
Sorry I thought I had it but I still don't understand it why does the expression in the parentheses approach 1? Take for example $frac{a_0}{x^{2n+1}}$ it is true if I take a big $x$ then the term goes as close to 0 as I want but I if I multiply this value by$ x^{2n+1}$ then I have$ a_0 $again which does not have to be $1$
$endgroup$
– RM777
Jan 11 at 20:05
$begingroup$
Yeah, typo on my part. The expression in parentheses approaches $a_{2n+1}$. And you have given the reason why: the other terms all go to zero.
$endgroup$
– Chris Custer
Jan 11 at 20:08
|
show 4 more comments
$begingroup$
Why can you Claim that so easily don't you have to give an explicit $a$ such that $forall_{rinmathbb{R}}exists_{ainmathbb{R}}forall_{xinmathbb{R}}:xin(a,+infty)Rightarrow p(x)>r$
$endgroup$
– RM777
Jan 11 at 19:32
$begingroup$
But no worries I could solve and understand the Problem now.
$endgroup$
– RM777
Jan 11 at 19:40
$begingroup$
Well that's fairly easy to do; given that the expression in parentheses approaches $1$. That is, take $x$ such that both $x^{2n+1}gtfrac r{a_{2n+1}}$ and the expression in parentheses is sufficiently close to $1$. There are some details to consider, but it clearly works out.
$endgroup$
– Chris Custer
Jan 11 at 19:45
$begingroup$
Sorry I thought I had it but I still don't understand it why does the expression in the parentheses approach 1? Take for example $frac{a_0}{x^{2n+1}}$ it is true if I take a big $x$ then the term goes as close to 0 as I want but I if I multiply this value by$ x^{2n+1}$ then I have$ a_0 $again which does not have to be $1$
$endgroup$
– RM777
Jan 11 at 20:05
$begingroup$
Yeah, typo on my part. The expression in parentheses approaches $a_{2n+1}$. And you have given the reason why: the other terms all go to zero.
$endgroup$
– Chris Custer
Jan 11 at 20:08
$begingroup$
Why can you Claim that so easily don't you have to give an explicit $a$ such that $forall_{rinmathbb{R}}exists_{ainmathbb{R}}forall_{xinmathbb{R}}:xin(a,+infty)Rightarrow p(x)>r$
$endgroup$
– RM777
Jan 11 at 19:32
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Why can you Claim that so easily don't you have to give an explicit $a$ such that $forall_{rinmathbb{R}}exists_{ainmathbb{R}}forall_{xinmathbb{R}}:xin(a,+infty)Rightarrow p(x)>r$
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– RM777
Jan 11 at 19:32
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But no worries I could solve and understand the Problem now.
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– RM777
Jan 11 at 19:40
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But no worries I could solve and understand the Problem now.
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– RM777
Jan 11 at 19:40
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Well that's fairly easy to do; given that the expression in parentheses approaches $1$. That is, take $x$ such that both $x^{2n+1}gtfrac r{a_{2n+1}}$ and the expression in parentheses is sufficiently close to $1$. There are some details to consider, but it clearly works out.
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– Chris Custer
Jan 11 at 19:45
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Well that's fairly easy to do; given that the expression in parentheses approaches $1$. That is, take $x$ such that both $x^{2n+1}gtfrac r{a_{2n+1}}$ and the expression in parentheses is sufficiently close to $1$. There are some details to consider, but it clearly works out.
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– Chris Custer
Jan 11 at 19:45
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Sorry I thought I had it but I still don't understand it why does the expression in the parentheses approach 1? Take for example $frac{a_0}{x^{2n+1}}$ it is true if I take a big $x$ then the term goes as close to 0 as I want but I if I multiply this value by$ x^{2n+1}$ then I have$ a_0 $again which does not have to be $1$
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– RM777
Jan 11 at 20:05
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Sorry I thought I had it but I still don't understand it why does the expression in the parentheses approach 1? Take for example $frac{a_0}{x^{2n+1}}$ it is true if I take a big $x$ then the term goes as close to 0 as I want but I if I multiply this value by$ x^{2n+1}$ then I have$ a_0 $again which does not have to be $1$
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– RM777
Jan 11 at 20:05
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Yeah, typo on my part. The expression in parentheses approaches $a_{2n+1}$. And you have given the reason why: the other terms all go to zero.
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– Chris Custer
Jan 11 at 20:08
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Yeah, typo on my part. The expression in parentheses approaches $a_{2n+1}$. And you have given the reason why: the other terms all go to zero.
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– Chris Custer
Jan 11 at 20:08
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You want to use this to prove the intermediate value theorem?
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– Mindlack
Jan 11 at 16:05
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Your statement is imprecise. The polynomial $P(x)=x$ has odd degree $1$ but does not have a positive or a negative zero.
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– MPW
Jan 11 at 16:05
1
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The claim in your title is not true. What are you actually trying to prove?
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– user3482749
Jan 11 at 16:10
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No what I want to show is that every polynomial of uneven Degree has a zeropoint. If I can show that for every such polynomial there exists a positive and a negative value i.e. $exists_{z,z'inmathbb{R}}:P(z)>0,P(z')<0$ I can make use of the intemediate value Theorem to say that $exists x$ in the intervall $[z,z']$ or $([z',z])$: $P(x)=0$. Therefor I assume without loss of generality that $P(x)>0,forall xin mathbb{R}$ and try to invoke a contradiction by constructing a $x$ with negative value.
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– RM777
Jan 11 at 16:11
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look at the limits. (if $P in mathbb{R}[X]$) it should give you the answer
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– Alexis
Jan 11 at 16:15