Derivative - Revision history

Lfox at 02:01, 2 July 2024

2024-07-02T02:01:28Z

← Older revision		Revision as of 02:01, 2 July 2024
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			The '''derivative''' <math>f'(x)</math>, of a differentiable function <math>f(x)</math>, is the slope of the tangent line of <math>f</math> at <math>x</math>. [TODO]

			Note that this definition does not talk about taking a limit as epsilon goes to 0.

			For some functions (indeed, many functions), there is a way to write <math>f'(x)</math> in terms of other known functions. That is, there is a relationship between the concept <math>f'(x)</math>, and other concepts. This need not be the case in general.

			== Old stuff [TODO delete] ==
	Let <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> be a [[function]], and let <math>\epsilon : \mathbb{Q}_{>0}</math>. I define <math>\Delta_\epsilon f : \mathbb{Q} \rightarrow \mathbb{Q}</math> by<math display="block">(\Delta_\epsilon f) (x) := \frac{f(x + \epsilon) - f(x)}{\epsilon}, \quad x : \mathbb{Q}. </math>I say that <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is '''differentiable''' at <math>x:\mathbb{Q}</math>, if <math>(\Delta_\epsilon f)(x) - (\Delta_{\epsilon'} f)(x) </math> is [[nill]] whenever <math>\epsilon, \epsilon' : \mathbb{Q}_{>0} </math> are both nill.		Let <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> be a [[function]], and let <math>\epsilon : \mathbb{Q}_{>0}</math>. I define <math>\Delta_\epsilon f : \mathbb{Q} \rightarrow \mathbb{Q}</math> by<math display="block">(\Delta_\epsilon f) (x) := \frac{f(x + \epsilon) - f(x)}{\epsilon}, \quad x : \mathbb{Q}. </math>I say that <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is '''differentiable''' at <math>x:\mathbb{Q}</math>, if <math>(\Delta_\epsilon f)(x) - (\Delta_{\epsilon'} f)(x) </math> is [[nill]] whenever <math>\epsilon, \epsilon' : \mathbb{Q}_{>0} </math> are both nill.

	If <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is differentiable at <math>x:\mathbb{Q}</math>, then I define the '''derivative''' of <math>f</math> at <math>x</math> as <math>f'(x) := (\Delta_\epsilon f)(x) </math>, for some nill <math>\epsilon : \mathbb{Q}_{>0}</math>.		If <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is differentiable at <math>x:\mathbb{Q}</math>, then I define the '''derivative''' of <math>f</math> at <math>x</math> as <math>f'(x) := (\Delta_\epsilon f)(x) </math>, for some nill <math>\epsilon : \mathbb{Q}_{>0}</math>.

	== Criticism ==		=== Criticism ===
	This definition makes things way more complicated. I will demonstrate this with the following example.		This definition makes things way more complicated. I will demonstrate this with the following example.

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	More generally, if <math>\nu : \mathbb{Q}_{>0}</math> is the nill cutoff (where we also assume <math>\nu \leq 1</math>), then <math>f(x) = x^3</math> is only "differentiable" in the region <math display="block">-\frac{1}{3}(1 + \nu) < x < \frac{1}{3}(1 - \nu)</math>This seems like an absurd conclusion.		More generally, if <math>\nu : \mathbb{Q}_{>0}</math> is the nill cutoff (where we also assume <math>\nu \leq 1</math>), then <math>f(x) = x^3</math> is only "differentiable" in the region <math display="block">-\frac{1}{3}(1 + \nu) < x < \frac{1}{3}(1 - \nu)</math>This seems like an absurd conclusion.

	== Polynomial derivatives ==		=== Polynomial derivatives ===
	Somewhat interesting, but also somewhat obvious vid https://www.youtube.com/watch?v=oW4jM0smS_E		Somewhat interesting, but also somewhat obvious vid https://www.youtube.com/watch?v=oW4jM0smS_E

	You can get the tangent line to a polynomial without actually invoking limits. My guess is that you can also get the tangent line to a rational function somehow.		You can get the tangent line to a polynomial without actually invoking limits. My guess is that you can also get the tangent line to a rational function somehow.

Lfox at 05:11, 23 April 2024

2024-04-23T05:11:43Z

← Older revision		Revision as of 05:11, 23 April 2024
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	== Polynomial derivatives ==		== Polynomial derivatives ==
	~~Interesting~~ vid https://www.youtube.com/watch?v=oW4jM0smS_E		Somewhat interesting, but also somewhat obvious vid https://www.youtube.com/watch?v=oW4jM0smS_E

			You can get the tangent line to a polynomial without actually invoking limits. My guess is that you can also get the tangent line to a rational function somehow.

Lfox at 04:25, 23 April 2024

2024-04-23T04:25:50Z

← Older revision		Revision as of 04:25, 23 April 2024
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	More generally, if <math>\nu : \mathbb{Q}_{>0}</math> is the nill cutoff (where we also assume <math>\nu \leq 1</math>), then <math>f(x) = x^3</math> is only "differentiable" in the region <math display="block">-\frac{1}{3}(1 + \nu) < x < \frac{1}{3}(1 - \nu)</math>This seems like an absurd conclusion.		More generally, if <math>\nu : \mathbb{Q}_{>0}</math> is the nill cutoff (where we also assume <math>\nu \leq 1</math>), then <math>f(x) = x^3</math> is only "differentiable" in the region <math display="block">-\frac{1}{3}(1 + \nu) < x < \frac{1}{3}(1 - \nu)</math>This seems like an absurd conclusion.

			== Polynomial derivatives ==
			Interesting vid https://www.youtube.com/watch?v=oW4jM0smS_E

Lfox: /* Criticism */

2024-01-28T22:58:43Z

Criticism

← Older revision		Revision as of 22:58, 28 January 2024
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	== Criticism ==		== Criticism ==
	This definition makes things way more complicated. I will demonstrate this with the following example. Let's suppose that <math>f(x) = x^3</math> so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^2 - \epsilon'^2)</math>. And let's say that we are in a context where anything with absolute value below 0.1 is nill. Let <math>\epsilon = 0.091 </math> and <math>\epsilon' = 0.001</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 0.27 x + 0.00828 </math>. This quantity is greater than 0.1, and thus non-nill, when <math>x > 0.339\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.339\overline{703}</math>.		This definition makes things way more complicated. I will demonstrate this with the following example.

	~~This complication is not present in the standard definition of derivative, where the remainder terms go away in the limit~~ and ~~we are just left with~~ <math>f'(x) = 3 x^2 </math>. It's ~~not completely clear whether or not this complication~~ is ~~a problem~~. ~~Those pesky terms are measuring something real~~, ~~which calculus~~ is ~~ignoring~~. ~~They are measuring~~ the ~~difference between two different methods of finding~~ the ~~slope of the tangent line to a real curve~~.		Let's suppose that <math>f(x) = x^3</math> so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^2 - \epsilon'^2)</math>. And let's say that we are in a context where anything with absolute value below 0.1 is nill. Let <math>\epsilon = 0.091 </math> and <math>\epsilon' = 0.001</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 0.27 x + 0.00828 </math>. This quantity is greater than 0.1, and thus non-nill, when <math>x > 0.339\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.339\overline{703}</math>.

			More generally, if <math>\nu : \mathbb{Q}_{>0}</math> is the nill cutoff (where we also assume <math>\nu \leq 1</math>), then <math>f(x) = x^3</math> is only "differentiable" in the region <math display="block">-\frac{1}{3}(1 + \nu) < x < \frac{1}{3}(1 - \nu)</math>This seems like an absurd conclusion.

Lfox at 22:17, 28 January 2024

2024-01-28T22:17:56Z

← Older revision		Revision as of 22:17, 28 January 2024
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	Let <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math>, and let <math>\epsilon : \mathbb{Q}_{>0}</math>. I define <math>\Delta_\epsilon f : \mathbb{Q} \rightarrow \mathbb{Q}</math> by<math display="block">(\Delta_\epsilon f) (x) := \frac{f(x + \epsilon) - f(x)}{\epsilon}, \quad x : \mathbb{Q}. </math>I say that <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is '''differentiable''' at <math>x:\mathbb{Q}</math>, if <math>(\Delta_\epsilon f)(x) - (\Delta_{\epsilon'} f)(x) </math> is [[nill]] whenever <math>\epsilon, \epsilon' : \mathbb{Q}_{>0} </math> are both nill.		Let <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> be a [[function]], and let <math>\epsilon : \mathbb{Q}_{>0}</math>. I define <math>\Delta_\epsilon f : \mathbb{Q} \rightarrow \mathbb{Q}</math> by<math display="block">(\Delta_\epsilon f) (x) := \frac{f(x + \epsilon) - f(x)}{\epsilon}, \quad x : \mathbb{Q}. </math>I say that <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is '''differentiable''' at <math>x:\mathbb{Q}</math>, if <math>(\Delta_\epsilon f)(x) - (\Delta_{\epsilon'} f)(x) </math> is [[nill]] whenever <math>\epsilon, \epsilon' : \mathbb{Q}_{>0} </math> are both nill.

	If <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is differentiable at <math>x:\mathbb{Q}</math>, then I define the '''derivative''' of <math>f</math> at <math>x</math> as <math>f'(x) := (\Delta_\epsilon f)(x) </math>, for some nill <math>\epsilon : \mathbb{Q}_{>0}</math>.		If <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is differentiable at <math>x:\mathbb{Q}</math>, then I define the '''derivative''' of <math>f</math> at <math>x</math> as <math>f'(x) := (\Delta_\epsilon f)(x) </math>, for some nill <math>\epsilon : \mathbb{Q}_{>0}</math>.

Lfox: /* Criticism */

2024-01-27T23:24:52Z

Criticism

← Older revision		Revision as of 23:24, 27 January 2024
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	This definition makes things way more complicated. I will demonstrate this with the following example. Let's suppose that <math>f(x) = x^3</math> so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^2 - \epsilon'^2)</math>. And let's say that we are in a context where anything with absolute value below 0.1 is nill. Let <math>\epsilon = 0.091 </math> and <math>\epsilon' = 0.001</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 0.27 x + 0.00828 </math>. This quantity is greater than 0.1, and thus non-nill, when <math>x > 0.339\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.339\overline{703}</math>.		This definition makes things way more complicated. I will demonstrate this with the following example. Let's suppose that <math>f(x) = x^3</math> so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^2 - \epsilon'^2)</math>. And let's say that we are in a context where anything with absolute value below 0.1 is nill. Let <math>\epsilon = 0.091 </math> and <math>\epsilon' = 0.001</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 0.27 x + 0.00828 </math>. This quantity is greater than 0.1, and thus non-nill, when <math>x > 0.339\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.339\overline{703}</math>.

	This complication is not present in the standard definition of derivative, where the remainder terms go away in the limit and we are just left with <math>f'(x) = 3 x^2 </math>. It's not completely clear whether or not this complication is a problem.		This complication is not present in the standard definition of derivative, where the remainder terms go away in the limit and we are just left with <math>f'(x) = 3 x^2 </math>. It's not completely clear whether or not this complication is a problem. Those pesky terms are measuring something real, which calculus is ignoring. They are measuring the difference between two different methods of finding the slope of the tangent line to a real curve.

Lfox: /* Criticism */

2024-01-27T02:33:02Z

Criticism

← Older revision		Revision as of 02:33, 27 January 2024
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	== Criticism ==		== Criticism ==
	This definition makes things way more complicated. I will demonstrate this with the following example. Let's suppose that <math>f(x) = x^3</math>, so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^3 - \epsilon'^3)</math>. And let's say that we are in a context where anything with absolute value below 1.0 is nill. Let <math>\epsilon = 0.91 </math> and <math>\epsilon' = 0.01</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 2.7 x + 0.~~75357~~</math>. This is greater than 1 when <math>x > 0.~~0912~~\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.~~0912~~\overline{703}</math>.		This definition makes things way more complicated. I will demonstrate this with the following example. Let's suppose that <math>f(x) = x^3</math> so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^2 - \epsilon'^2)</math>. And let's say that we are in a context where anything with absolute value below 0.1 is nill. Let <math>\epsilon = 0.091 </math> and <math>\epsilon' = 0.001</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 0.27 x + 0.00828 </math>. This quantity is greater than 0.1, and thus non-nill, when <math>x > 0.339\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.339\overline{703}</math>.

	This complication is not present in the standard definition of derivative, where the remainder terms go away in the limit and we are just left with <math>f'(x) = 3 x^2 </math>. It's not completely clear whether or not this complication is a problem.		This complication is not present in the standard definition of derivative, where the remainder terms go away in the limit and we are just left with <math>f'(x) = 3 x^2 </math>. It's not completely clear whether or not this complication is a problem.

Lfox: /* Criticism */

2024-01-27T02:14:55Z

Criticism

← Older revision		Revision as of 02:14, 27 January 2024
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	This definition makes things way more complicated. I will demonstrate this with the following example. Let's suppose that <math>f(x) = x^3</math>, so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^3 - \epsilon'^3)</math>. And let's say that we are in a context where anything with absolute value below 1.0 is nill. Let <math>\epsilon = 0.91 </math> and <math>\epsilon' = 0.01</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 2.7 x + 0.75357</math>. This is greater than 1 when <math>x > 0.0912\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.0912\overline{703}</math>.		This definition makes things way more complicated. I will demonstrate this with the following example. Let's suppose that <math>f(x) = x^3</math>, so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^3 - \epsilon'^3)</math>. And let's say that we are in a context where anything with absolute value below 1.0 is nill. Let <math>\epsilon = 0.91 </math> and <math>\epsilon' = 0.01</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 2.7 x + 0.75357</math>. This is greater than 1 when <math>x > 0.0912\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.0912\overline{703}</math>.

	This ~~problem~~ is not present in the standard definition of derivative, where the remainder terms go away in the limit and we are just left with <math>f'(x) = 3 x^2 </math>.		This complication is not present in the standard definition of derivative, where the remainder terms go away in the limit and we are just left with <math>f'(x) = 3 x^2 </math>. It's not completely clear whether or not this complication is a problem.

Lfox at 21:55, 26 January 2024

2024-01-26T21:55:38Z

← Older revision		Revision as of 21:55, 26 January 2024
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	If <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is differentiable at <math>x:\mathbb{Q}</math>, then I define the '''derivative''' of <math>f</math> at <math>x</math> as <math>f'(x) := (\Delta_\epsilon f)(x) </math>, for some nill <math>\epsilon : \mathbb{Q}_{>0}</math>.		If <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is differentiable at <math>x:\mathbb{Q}</math>, then I define the '''derivative''' of <math>f</math> at <math>x</math> as <math>f'(x) := (\Delta_\epsilon f)(x) </math>, for some nill <math>\epsilon : \mathbb{Q}_{>0}</math>.

			== Criticism ==
			This definition makes things way more complicated. I will demonstrate this with the following example. Let's suppose that <math>f(x) = x^3</math>, so <math>(\Delta_\epsilon f)(x) = 3x^2 + 3\epsilon x + \epsilon^2 </math> and <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 3 (\epsilon - \epsilon' )x + ( \epsilon^3 - \epsilon'^3)</math>. And let's say that we are in a context where anything with absolute value below 1.0 is nill. Let <math>\epsilon = 0.91 </math> and <math>\epsilon' = 0.01</math>. Then <math>(\Delta_\epsilon f - \Delta_{\epsilon'} f)(x) = 2.7 x + 0.75357</math>. This is greater than 1 when <math>x > 0.0912\overline{703}</math>, so we reach the seemingly absurd conclusion that <math>f(x) = x^3</math> is not differentiable where <math>x > 0.0912\overline{703}</math>.

			This problem is not present in the standard definition of derivative, where the remainder terms go away in the limit and we are just left with <math>f'(x) = 3 x^2 </math>.

Lfox at 20:25, 26 January 2024

2024-01-26T20:25:20Z

← Older revision		Revision as of 20:25, 26 January 2024
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	Let <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math>, and let <math>\epsilon : \mathbb{Q}</math> ~~be positive~~. We define <math>\Delta_\epsilon f : \mathbb{Q} \rightarrow \mathbb{Q}</math> by<math display="block">(\Delta_\epsilon f) (x) := \frac{f(x + \epsilon) - f(x)}{\epsilon}. </math>We say that <math>f</math> is '''differentiable''' if is ~~....~~ [[nill]] whenever		Let <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math>, and let <math>\epsilon : \mathbb{Q}_{>0}</math>. I define <math>\Delta_\epsilon f : \mathbb{Q} \rightarrow \mathbb{Q}</math> by<math display="block">(\Delta_\epsilon f) (x) := \frac{f(x + \epsilon) - f(x)}{\epsilon}, \quad x : \mathbb{Q}. </math>I say that <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is '''differentiable''' at <math>x:\mathbb{Q}</math>, if <math>(\Delta_\epsilon f)(x) - (\Delta_{\epsilon'} f)(x) </math> is [[nill]] whenever <math>\epsilon, \epsilon' : \mathbb{Q}_{>0} </math> are both nill.

			If <math>f : \mathbb{Q} \rightarrow \mathbb{Q}</math> is differentiable at <math>x:\mathbb{Q}</math>, then I define the '''derivative''' of <math>f</math> at <math>x</math> as <math>f'(x) := (\Delta_\epsilon f)(x) </math>, for some nill <math>\epsilon : \mathbb{Q}_{>0}</math>.