Mixing
Is my proof of second fundamental theorem of calculus (without mean value theorem) valid?
$begingroup$
Second Fundamental Theorem Of Calculus (The Evaluation Theorem):
For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:
$$int_A^B f(x)dx = F(B) - F(A)$$
Let $P = {x_i}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:
$$sum_{i=1}^n f(x_i)Delta x$$
By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:
$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = int_A^B f(x)dx$$
Where
$$Delta x = frac{B - A}{n}$$
$$x_i = A + i*Delta x$$
and so that
$$x_0 = A$$
$$x_n = A + n*frac{B - A}{n} = A + B - A = B.$$
Since we also know that
$$n to infty Rightarrow Delta x to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = lim_{Delta x to 0} frac{F(x_i) - F(x_i - Delta x)}{Delta x},$$
we can write that
$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} f(x_i)Delta x.$$
And following that we can notate the series without need of a second limit via applying the limit of $Delta x$ to each element of the sum (other than this, it only affects the total count of elements. Also look to the clarificiation at the bottom of the question) as
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x),$$
which is that
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x) = lim_{n to infty}sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$
And so what we obtain is equivalent with
$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$
which is nothing but
$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$
Following that:
$$int_A^B f(x)dx = F(B) - F(A)$$
Q.E.D.
Clarification for the part "without need of a second limit via applying the limit of $Delta x$ to each element of sum":
$$lim_{Delta x to 0} f(x_i)Delta x = lim_{Delta x to 0} (f(x_i)) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x}) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x} * Delta x). $$
So we can continue with
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x. $$
calculus
asked Jan 18 at 16:22
İbrahim İpekİbrahim İpek
197
$endgroup$
|
show 11 more comments
$begingroup$
Second Fundamental Theorem Of Calculus (The Evaluation Theorem):
For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:
$$int_A^B f(x)dx = F(B) - F(A)$$
Let $P = {x_i}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:
$$sum_{i=1}^n f(x_i)Delta x$$
By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:
$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = int_A^B f(x)dx$$
Where
$$Delta x = frac{B - A}{n}$$
$$x_i = A + i*Delta x$$
and so that
$$x_0 = A$$
$$x_n = A + n*frac{B - A}{n} = A + B - A = B.$$
Since we also know that
$$n to infty Rightarrow Delta x to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = lim_{Delta x to 0} frac{F(x_i) - F(x_i - Delta x)}{Delta x},$$
we can write that
$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} f(x_i)Delta x.$$
And following that we can notate the series without need of a second limit via applying the limit of $Delta x$ to each element of the sum (other than this, it only affects the total count of elements. Also look to the clarificiation at the bottom of the question) as
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x),$$
which is that
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x) = lim_{n to infty}sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$
And so what we obtain is equivalent with
$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$
which is nothing but
$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$
Following that:
$$int_A^B f(x)dx = F(B) - F(A)$$
Q.E.D.
Clarification for the part "without need of a second limit via applying the limit of $Delta x$ to each element of sum":
$$lim_{Delta x to 0} f(x_i)Delta x = lim_{Delta x to 0} (f(x_i)) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x}) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x} * Delta x). $$
So we can continue with
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x. $$
calculus
asked Jan 18 at 16:22
İbrahim İpekİbrahim İpek
197
$endgroup$
1
$begingroup$
In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
$endgroup$
– Randall
Jan 18 at 16:24
1
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$F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
$endgroup$
– İbrahim İpek
Jan 18 at 17:05
1
$begingroup$
I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
$endgroup$
– Cardioid_Ass_22
Jan 18 at 17:30
1
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@İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
$endgroup$
– Arturo Magidin
Jan 18 at 20:47
3
$begingroup$
Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
$endgroup$
– Arturo Magidin
Jan 18 at 20:52
|
show 11 more comments
$begingroup$
Second Fundamental Theorem Of Calculus (The Evaluation Theorem):
For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:
$$int_A^B f(x)dx = F(B) - F(A)$$
Let $P = {x_i}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:
$$sum_{i=1}^n f(x_i)Delta x$$
By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:
$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = int_A^B f(x)dx$$
Where
$$Delta x = frac{B - A}{n}$$
$$x_i = A + i*Delta x$$
and so that
$$x_0 = A$$
$$x_n = A + n*frac{B - A}{n} = A + B - A = B.$$
Since we also know that
$$n to infty Rightarrow Delta x to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = lim_{Delta x to 0} frac{F(x_i) - F(x_i - Delta x)}{Delta x},$$
we can write that
$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} f(x_i)Delta x.$$
And following that we can notate the series without need of a second limit via applying the limit of $Delta x$ to each element of the sum (other than this, it only affects the total count of elements. Also look to the clarificiation at the bottom of the question) as
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x),$$
which is that
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x) = lim_{n to infty}sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$
And so what we obtain is equivalent with
$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$
which is nothing but
$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$
Following that:
$$int_A^B f(x)dx = F(B) - F(A)$$
Q.E.D.
Clarification for the part "without need of a second limit via applying the limit of $Delta x$ to each element of sum":
$$lim_{Delta x to 0} f(x_i)Delta x = lim_{Delta x to 0} (f(x_i)) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x}) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x} * Delta x). $$
So we can continue with
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x. $$
calculus
asked Jan 18 at 16:22
İbrahim İpekİbrahim İpek
197
$endgroup$
Second Fundamental Theorem Of Calculus (The Evaluation Theorem):
For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:
$$int_A^B f(x)dx = F(B) - F(A)$$
Let $P = {x_i}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:
$$sum_{i=1}^n f(x_i)Delta x$$
By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:
$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = int_A^B f(x)dx$$
Where
$$Delta x = frac{B - A}{n}$$
$$x_i = A + i*Delta x$$
and so that
$$x_0 = A$$
$$x_n = A + n*frac{B - A}{n} = A + B - A = B.$$
Since we also know that
$$n to infty Rightarrow Delta x to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = lim_{Delta x to 0} frac{F(x_i) - F(x_i - Delta x)}{Delta x},$$
we can write that
$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} f(x_i)Delta x.$$
And following that we can notate the series without need of a second limit via applying the limit of $Delta x$ to each element of the sum (other than this, it only affects the total count of elements. Also look to the clarificiation at the bottom of the question) as
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x),$$
which is that
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x) = lim_{n to infty}sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$
And so what we obtain is equivalent with
$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$
which is nothing but
$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$
Following that:
$$int_A^B f(x)dx = F(B) - F(A)$$
Q.E.D.
Clarification for the part "without need of a second limit via applying the limit of $Delta x$ to each element of sum":
$$lim_{Delta x to 0} f(x_i)Delta x = lim_{Delta x to 0} (f(x_i)) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x}) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x} * Delta x). $$
So we can continue with
$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x. $$
calculus
calculus
asked Jan 18 at 16:22
İbrahim İpekİbrahim İpek
197
asked Jan 18 at 16:22
İbrahim İpekİbrahim İpek
197
edited Jan 18 at 20:39
İbrahim İpek
asked Jan 18 at 16:22
İbrahim İpekİbrahim İpek
197
asked Jan 18 at 16:22
İbrahim İpekİbrahim İpek
197
asked Jan 18 at 16:22
İbrahim İpekİbrahim İpek
197
197
1
$begingroup$
In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
$endgroup$
– Randall
Jan 18 at 16:24
1
$begingroup$
$F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
$endgroup$
– İbrahim İpek
Jan 18 at 17:05
1
$begingroup$
I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
$endgroup$
– Cardioid_Ass_22
Jan 18 at 17:30
1
$begingroup$
@İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
$endgroup$
– Arturo Magidin
Jan 18 at 20:47
3
$begingroup$
Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
$endgroup$
– Arturo Magidin
Jan 18 at 20:52
|
show 11 more comments
1
$begingroup$
In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
$endgroup$
– Randall
Jan 18 at 16:24
1
$begingroup$
$F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
$endgroup$
– İbrahim İpek
Jan 18 at 17:05
1
$begingroup$
I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
$endgroup$
– Cardioid_Ass_22
Jan 18 at 17:30
1
$begingroup$
@İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
$endgroup$
– Arturo Magidin
Jan 18 at 20:47
3
$begingroup$
Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
$endgroup$
– Arturo Magidin
Jan 18 at 20:52
1
1
$begingroup$
In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
$endgroup$
– Randall
Jan 18 at 16:24
$begingroup$
In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
$endgroup$
– Randall
Jan 18 at 16:24
1
1
$begingroup$
$F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
$endgroup$
– İbrahim İpek
Jan 18 at 17:05
$begingroup$
$F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
$endgroup$
– İbrahim İpek
Jan 18 at 17:05
1
1
$begingroup$
I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
$endgroup$
– Cardioid_Ass_22
Jan 18 at 17:30
$begingroup$
I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
$endgroup$
– Cardioid_Ass_22
Jan 18 at 17:30
1
1
$begingroup$
@İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
$endgroup$
– Arturo Magidin
Jan 18 at 20:47
$begingroup$
@İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
$endgroup$
– Arturo Magidin
Jan 18 at 20:47
3
3
$begingroup$
Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
$endgroup$
– Arturo Magidin
Jan 18 at 20:52
$begingroup$
Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
$endgroup$
– Arturo Magidin
Jan 18 at 20:52
|
show 11 more comments
2 Answers
2
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votes
$begingroup$
Your proof would be correct, if you can first prove something like the following:
Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
$$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$
For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
As an indication that this isn't easy, note that it isn't true without additional assumptions:
$$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$
answered Jan 18 at 21:44
Calvin KhorCalvin Khor
12.1k21438
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add a comment |
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I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.
Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.
First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.
Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.
By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$
It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.
answered Jan 19 at 3:22
Paramanand SinghParamanand Singh
50.4k556166
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add a comment |
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$begingroup$
Your proof would be correct, if you can first prove something like the following:
Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
$$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$
For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
As an indication that this isn't easy, note that it isn't true without additional assumptions:
$$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$
answered Jan 18 at 21:44
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
add a comment |
$begingroup$
Your proof would be correct, if you can first prove something like the following:
Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
$$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$
For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
As an indication that this isn't easy, note that it isn't true without additional assumptions:
$$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$
answered Jan 18 at 21:44
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
add a comment |
$begingroup$
Your proof would be correct, if you can first prove something like the following:
Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
$$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$
For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
As an indication that this isn't easy, note that it isn't true without additional assumptions:
$$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$
answered Jan 18 at 21:44
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
Your proof would be correct, if you can first prove something like the following:
Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
$$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$
For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
As an indication that this isn't easy, note that it isn't true without additional assumptions:
$$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$
answered Jan 18 at 21:44
Calvin KhorCalvin Khor
12.1k21438
edited Jan 18 at 21:50
answered Jan 18 at 21:44
Calvin KhorCalvin Khor
12.1k21438
answered Jan 18 at 21:44
Calvin KhorCalvin Khor
12.1k21438
answered Jan 18 at 21:44
Calvin KhorCalvin Khor
12.1k21438
12.1k21438
add a comment |
add a comment |
$begingroup$
I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.
Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.
First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.
Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.
By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$
It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.
answered Jan 19 at 3:22
Paramanand SinghParamanand Singh
50.4k556166
$endgroup$
add a comment |
$begingroup$
I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.
Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.
First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.
Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.
By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$
It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.
answered Jan 19 at 3:22
Paramanand SinghParamanand Singh
50.4k556166
$endgroup$
add a comment |
$begingroup$
I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.
Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.
First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.
Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.
By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$
It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.
answered Jan 19 at 3:22
Paramanand SinghParamanand Singh
50.4k556166
$endgroup$
I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.
Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.
First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.
Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.
By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$
It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.
answered Jan 19 at 3:22
Paramanand SinghParamanand Singh
50.4k556166
answered Jan 19 at 3:22
Paramanand SinghParamanand Singh
50.4k556166
answered Jan 19 at 3:22
Paramanand SinghParamanand Singh
50.4k556166
answered Jan 19 at 3:22
Paramanand SinghParamanand Singh
50.4k556166
50.4k556166
add a comment |
add a comment |
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1
$begingroup$
In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
$endgroup$
– Randall
Jan 18 at 16:24
1
$begingroup$
$F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
$endgroup$
– İbrahim İpek
Jan 18 at 17:05
1
$begingroup$
I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
$endgroup$
– Cardioid_Ass_22
Jan 18 at 17:30
1
$begingroup$
@İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
$endgroup$
– Arturo Magidin
Jan 18 at 20:47
3
$begingroup$
Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
$endgroup$
– Arturo Magidin
Jan 18 at 20:52