Show that if $P(x)$ is an polynomial with an uneven degree then there exists a $x_0in mathbb{R}: P(x_0) <...












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 16:19







RM777

















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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















2












$begingroup$

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, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.






share|cite|improve this answer









$endgroup$



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, $lim_{xtopminfty}p(x)=lim_{xtopminfty}x^{2n+1}(a_{2n+1}+frac{a_{2n}}{x}+frac{a_{2n-1}}{x^2} +dots+frac{a_0}{x^{2n+1}})=lim_{ntopminfty}a_{2n+1}x^{2n+1}=pminfty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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






$begingroup$
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




$begingroup$
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






$begingroup$
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




$begingroup$
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




$begingroup$
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


















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