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 1 hour ago
Martin Sleziak
44.7k8115271
asked 1 hour ago
Steve
426
|
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 1 hour ago
Martin Sleziak
44.7k8115271
asked 1 hour ago
Steve
426
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
1 hour ago
Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
– Steve
1 hour ago
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
1 hour ago
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
55 mins ago
1
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
53 mins ago
|
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 1 hour ago
Martin Sleziak
44.7k8115271
asked 1 hour ago
Steve
426
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 1 hour ago
Martin Sleziak
44.7k8115271
asked 1 hour ago
Steve
426
edited 1 hour ago
Martin Sleziak
44.7k8115271
asked 1 hour ago
Steve
426
edited 1 hour ago
Martin Sleziak
44.7k8115271
edited 1 hour ago
Martin Sleziak
44.7k8115271
edited 1 hour ago
Martin Sleziak
44.7k8115271
44.7k8115271
asked 1 hour ago
Steve
426
asked 1 hour ago
Steve
426
asked 1 hour ago
Steve
426
426
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
1 hour ago
Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
– Steve
1 hour ago
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
1 hour ago
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
55 mins ago
1
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
53 mins ago
|
show 6 more comments
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
1 hour ago
Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
– Steve
1 hour ago
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
1 hour ago
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
55 mins ago
1
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
53 mins ago
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
1 hour ago
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
1 hour ago
Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
– Steve
1 hour ago
Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
– Steve
1 hour ago
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
1 hour ago
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
1 hour ago
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
55 mins ago
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
55 mins ago
1
1
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
53 mins ago
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
53 mins ago
|
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
$displaystyle lim_{n mathop to infty}f(x_n)g(x_n) = f(displaystyle lim_{n mathop to infty} x_n )g(displaystyle lim_{n mathop to infty} x_n) = f(a)g(a)$
answered 1 hour ago
gandalf61
7,781623
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 1 hour ago
Todor Markov
1,55839
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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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
$displaystyle lim_{n mathop to infty}f(x_n)g(x_n) = f(displaystyle lim_{n mathop to infty} x_n )g(displaystyle lim_{n mathop to infty} x_n) = f(a)g(a)$
answered 1 hour ago
gandalf61
7,781623
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
$displaystyle lim_{n mathop to infty}f(x_n)g(x_n) = f(displaystyle lim_{n mathop to infty} x_n )g(displaystyle lim_{n mathop to infty} x_n) = f(a)g(a)$
answered 1 hour ago
gandalf61
7,781623
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
$displaystyle lim_{n mathop to infty}f(x_n)g(x_n) = f(displaystyle lim_{n mathop to infty} x_n )g(displaystyle lim_{n mathop to infty} x_n) = f(a)g(a)$
answered 1 hour ago
gandalf61
7,781623
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
$displaystyle lim_{n mathop to infty}f(x_n)g(x_n) = f(displaystyle lim_{n mathop to infty} x_n )g(displaystyle lim_{n mathop to infty} x_n) = f(a)g(a)$
answered 1 hour ago
gandalf61
7,781623
answered 1 hour ago
gandalf61
7,781623
answered 1 hour ago
gandalf61
7,781623
answered 1 hour ago
gandalf61
7,781623
7,781623
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 1 hour ago
Todor Markov
1,55839
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 1 hour ago
Todor Markov
1,55839
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 1 hour ago
Todor Markov
1,55839
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 1 hour ago
Todor Markov
1,55839
answered 1 hour ago
Todor Markov
1,55839
answered 1 hour ago
Todor Markov
1,55839
answered 1 hour ago
Todor Markov
1,55839
1,55839
add a comment |
add a comment |
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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
1 hour ago
Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
– Steve
1 hour ago
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
1 hour ago
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
55 mins ago
1
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
53 mins ago