Need Help : Proving polynomials are continuous, without circular reasoning












12














I know there are a lot of answers regarding continuity of polynomials. But, this question is different.



We need to have $ lim_{xto a} {x^n} = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_{xto a} {f(x)} = L$ and $ lim_{xto a} {g(x)} = K$



Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrt{epsilon}$



And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrt{epsilon}$



Let $δ=min{δ_1,δ_2}$



Hence, $0<|x-a|<δ$



⟹$|(f(x)-L)(g(x)-K)-0|<sqrt{epsilon} sqrt{epsilon} = ϵ$



Hence, $lim_{x to a} {(f(x)-L)(g(x)-K)} = 0$



⟹$lim_{x to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$



And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.






real-analysis limits continuity epsilon-delta







share|cite|improve this question




















  • 1




    You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
    – Martin Sleziak
    yesterday












  • Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
    – Steve
    yesterday








  • 2




    Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
    – Paramanand Singh
    yesterday






  • 1




    Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
    – Paramanand Singh
    yesterday






  • 2




    In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
    – Paramanand Singh
    yesterday
















12














I know there are a lot of answers regarding continuity of polynomials. But, this question is different.



We need to have $ lim_{xto a} {x^n} = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_{xto a} {f(x)} = L$ and $ lim_{xto a} {g(x)} = K$



Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrt{epsilon}$



And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrt{epsilon}$



Let $δ=min{δ_1,δ_2}$



Hence, $0<|x-a|<δ$



⟹$|(f(x)-L)(g(x)-K)-0|<sqrt{epsilon} sqrt{epsilon} = ϵ$



Hence, $lim_{x to a} {(f(x)-L)(g(x)-K)} = 0$



⟹$lim_{x to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$



And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.






real-analysis limits continuity epsilon-delta







share|cite|improve this question




















  • 1




    You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
    – Martin Sleziak
    yesterday












  • Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
    – Steve
    yesterday








  • 2




    Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
    – Paramanand Singh
    yesterday






  • 1




    Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
    – Paramanand Singh
    yesterday






  • 2




    In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
    – Paramanand Singh
    yesterday














12












12








12


0





I know there are a lot of answers regarding continuity of polynomials. But, this question is different.



We need to have $ lim_{xto a} {x^n} = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_{xto a} {f(x)} = L$ and $ lim_{xto a} {g(x)} = K$



Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrt{epsilon}$



And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrt{epsilon}$



Let $δ=min{δ_1,δ_2}$



Hence, $0<|x-a|<δ$



⟹$|(f(x)-L)(g(x)-K)-0|<sqrt{epsilon} sqrt{epsilon} = ϵ$



Hence, $lim_{x to a} {(f(x)-L)(g(x)-K)} = 0$



⟹$lim_{x to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$



And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.






real-analysis limits continuity epsilon-delta







share|cite|improve this question















I know there are a lot of answers regarding continuity of polynomials. But, this question is different.



We need to have $ lim_{xto a} {x^n} = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_{xto a} {f(x)} = L$ and $ lim_{xto a} {g(x)} = K$



Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrt{epsilon}$



And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrt{epsilon}$



Let $δ=min{δ_1,δ_2}$



Hence, $0<|x-a|<δ$



⟹$|(f(x)-L)(g(x)-K)-0|<sqrt{epsilon} sqrt{epsilon} = ϵ$



Hence, $lim_{x to a} {(f(x)-L)(g(x)-K)} = 0$



⟹$lim_{x to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$



And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.






real-analysis limits continuity epsilon-delta




real-analysis limits continuity epsilon-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited yesterday









Martin Sleziak

44.6k8115271




44.6k8115271










asked yesterday









SteveSteve

828




828








  • 1




    You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
    – Martin Sleziak
    yesterday












  • Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
    – Steve
    yesterday








  • 2




    Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
    – Paramanand Singh
    yesterday






  • 1




    Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
    – Paramanand Singh
    yesterday






  • 2




    In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
    – Paramanand Singh
    yesterday














  • 1




    You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
    – Martin Sleziak
    yesterday












  • Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
    – Steve
    yesterday








  • 2




    Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
    – Paramanand Singh
    yesterday






  • 1




    Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
    – Paramanand Singh
    yesterday






  • 2




    In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
    – Paramanand Singh
    yesterday








1




1




You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
– Martin Sleziak
yesterday






You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
– Martin Sleziak
yesterday














Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
– Steve
yesterday






Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
– Steve
yesterday






2




2




Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
– Paramanand Singh
yesterday




Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
– Paramanand Singh
yesterday




1




1




Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
– Paramanand Singh
yesterday




Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
– Paramanand Singh
yesterday




2




2




In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
– Paramanand Singh
yesterday




In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
– Paramanand Singh
yesterday










2 Answers
2






active

oldest

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17














The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



$$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$






share|cite|improve this answer































    10














    You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



    For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.






    share|cite|improve this answer





















    • IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
      – leftaroundabout
      22 hours ago













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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

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    17














    The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



    $$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$






    share|cite|improve this answer




























      17














      The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



      $$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$






      share|cite|improve this answer


























        17












        17








        17






        The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



        $$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$






        share|cite|improve this answer














        The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



        $$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday









        Mutantoe

        576412




        576412










        answered yesterday









        gandalf61gandalf61

        7,931625




        7,931625























            10














            You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



            For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.






            share|cite|improve this answer





















            • IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
              – leftaroundabout
              22 hours ago


















            10














            You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



            For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.






            share|cite|improve this answer





















            • IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
              – leftaroundabout
              22 hours ago
















            10












            10








            10






            You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



            For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.






            share|cite|improve this answer












            You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



            For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            Todor MarkovTodor Markov

            1,664410




            1,664410












            • IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
              – leftaroundabout
              22 hours ago




















            • IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
              – leftaroundabout
              22 hours ago


















            IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
            – leftaroundabout
            22 hours ago






            IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
            – leftaroundabout
            22 hours ago




















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            An IMO inspired problem

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            14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
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            Management

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            Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
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            Has there ever been an instance of an active nuclear power plant within or near a war zone?

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            36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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