Need Help : Proving polynomials are continuous, without circular reasoning
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
edited yesterday
Martin Sleziak
44.6k8115271
asked yesterday
SteveSteve
828
|
show 6 more comments
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
edited yesterday
Martin Sleziak
44.6k8115271
asked yesterday
SteveSteve
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
|
show 6 more comments
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
edited yesterday
Martin Sleziak
44.6k8115271
asked yesterday
SteveSteve
828
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
edited yesterday
Martin Sleziak
44.6k8115271
asked yesterday
SteveSteve
828
edited yesterday
Martin Sleziak
44.6k8115271
asked yesterday
SteveSteve
828
edited yesterday
Martin Sleziak
44.6k8115271
edited yesterday
Martin Sleziak
44.6k8115271
edited yesterday
Martin Sleziak
44.6k8115271
44.6k8115271
asked yesterday
SteveSteve
828
asked yesterday
SteveSteve
828
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
|
show 6 more comments
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
|
show 6 more comments
2 Answers
2
active
oldest
votes
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
$$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$
edited yesterday
Mutantoe
572412
answered yesterday
gandalf61gandalf61
7,921624
add a comment |
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 $epsilon^prime$.
answered yesterday
Todor MarkovTodor Markov
1,619410
IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
– leftaroundabout
15 hours ago
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
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
$$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$
edited yesterday
Mutantoe
572412
answered yesterday
gandalf61gandalf61
7,921624
add a comment |
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
$$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$
edited yesterday
Mutantoe
572412
answered yesterday
gandalf61gandalf61
7,921624
add a comment |
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
$$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$
edited yesterday
Mutantoe
572412
answered yesterday
gandalf61gandalf61
7,921624
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
$$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$
edited yesterday
Mutantoe
572412
answered yesterday
gandalf61gandalf61
7,921624
edited yesterday
Mutantoe
572412
edited yesterday
Mutantoe
572412
edited yesterday
Mutantoe
572412
572412
answered yesterday
gandalf61gandalf61
7,921624
answered yesterday
gandalf61gandalf61
7,921624
answered yesterday
gandalf61gandalf61
7,921624
7,921624
add a comment |
add a comment |
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 $epsilon^prime$.
answered yesterday
Todor MarkovTodor Markov
1,619410
IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
– leftaroundabout
15 hours ago
add a comment |
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 $epsilon^prime$.
answered yesterday
Todor MarkovTodor Markov
1,619410
IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
– leftaroundabout
15 hours ago
add a comment |
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 $epsilon^prime$.
answered yesterday
Todor MarkovTodor Markov
1,619410
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 $epsilon^prime$.
answered yesterday
Todor MarkovTodor Markov
1,619410
answered yesterday
Todor MarkovTodor Markov
1,619410
answered yesterday
Todor MarkovTodor Markov
1,619410
answered yesterday
Todor MarkovTodor Markov
1,619410
1,619410
IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
– leftaroundabout
15 hours ago
add a comment |
IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
– leftaroundabout
15 hours ago
IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
– leftaroundabout
15 hours ago
IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
– leftaroundabout
15 hours ago
add a comment |
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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