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[SOLVED] relationship obtaining between various derivatives

DeusAbscondus

Active member
Jun 30, 2012
176
I'm just experimenting with graphs of derivatives and have noticed something
that our teacher has never adverted to but strikes me as very interesting:

the third derivative when set to zero will give the inflection points of the first, which makes me reflect that the relationship which obtains between, say, the 1st and third, must be mathematically equivalent to the relationship obtaining between the 2nd and fourth, for instance.

Is this humble conjecture clear? if so, is it true?

I've included a graph which tries to illustrate this (I am fully aware of how basic this must seem to you guys, but *I* am having fun!:D)

Thanks for any comments,
Deus Abs
 
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Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
the third derivative when set to zero will give the inflection points of the first
This is clear.

which makes me reflect that the relationship between each successive derivative and the function from which it is derived must be mathematically the same in each case.
This is not so clear. Of course, in a sequence of functions: f, f', f'', f''', ..., each successive function is a derivative or a previous function.

The fact you observed holds not just for derivatives. "If x is an inflection point for f then the second derivative, f″(x), is equal to zero if it exists" (Wikipedia).
 

DeusAbscondus

Active member
Jun 30, 2012
176
This is clear.

This is not so clear. Of course, in a sequence of functions: f, f', f'', f''', ..., each successive function is a derivative or a previous function.

The fact you observed holds not just for derivatives. "If x is an inflection point for f then the second derivative, f″(x), is equal to zero if it exists" (Wikipedia).
Thanks for pointing out the lack of clarity.
Do my italized words (just now edited) help clarify my point?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
the relationship which obtains between, say, the 1st and third, must be mathematically equivalent to the relationship obtaining between the 2nd and fourth, for instance.
If there is some relationship between f and f'' for all all functions f that are differentiable two times, then the same relationship exists between $f^{(n)}$ and $f^{(n+2)}$ where the derivatives exist. Here $f^{(n)}$ is the nth derivative. This is completely obvious because $f^{(n+2)}=\left(f^{(n)}\right)''$.
 

DeusAbscondus

Active member
Jun 30, 2012
176
If there is some relationship between f and f'' for all all functions f that are differentiable two times, then the same relationship exists between $f^{(n)}$ and $f^{(n+2)}$ where the derivatives exist. Here $f^{(n)}$ is the nth derivative. This is completely obvious because $f^{(n+2)}=\left(f^{(n)}\right)''$.
It may be "completely obvious" to you; to me it is anything but completely obvious as I am having to work things out pretty much by myself.

Thank you for confirming my tentative obversation (which was not offered as a mathematical breakthrough but a personal experience of a new learning, seeking confirmation)

Deus Abs
 

DeusAbscondus

Active member
Jun 30, 2012
176
$\text{What inference can be drawn for all} f(x) when f(x)\ \text{is twice-differentiable and}$
$f(x)\ \text{contains at least one inflection point, as in the case of:}$
$$f(x)=x^3-x^2+4x-3$$
$$f'(x)=3x^2-2x+4$$
$$f''(x)=6x-2$$

$\text{I have included a graph of above, showing perpendicular drawn through points A, B and C, being, respectively: }$
$A=\text{inflection point} (0.33, -1.74) on f(x)$
$B=\text{the intersection between} f''(x) \text{and x-axis, or, simply}f''(0), and$
$C=(0.33, 3.67), \text{being Minimum Turning Point on} f'(x)$
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
What kind of connection are you potentially seeing or think might be possible? As I follow your last post x=.33 is the point of inflection for f(x) and you're looking at f(.33), f'(.33) and f''(.33)? They are along a vertical line because we chose x=.33 for all of them so the fact that they are vertically aligned is not significant. Are you looking at the vertical spacing between the points?
 

DeusAbscondus

Active member
Jun 30, 2012
176
What kind of connection are you potentially seeing or think might be possible? As I follow your last post x=.33 is the point of inflection for f(x) and you're looking at f(.33), f'(.33) and f''(.33)? They are along a vertical line because we chose x=.33 for all of them so the fact that they are vertically aligned is not significant. Are you looking at the vertical spacing between the points?
Hi Jameson,
note appended to graph attached.
Regs
Deus Abs

Erratum:
sorry:
Read: "is it generally true that a function which has a point of inflection and which is twice differentiable..."
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Sorry to ask for more explaining but I'm just not getting something. What features do you mean here?

It this example you posted f(x), f'(x) and f'''(x) as graphs. You solved for f'(x)=0 which is at x = -.44. Then you found f(-.44) and f''(-.44) and plotted those points, and finally connected all three points with a dotted line.

Again, the reason they are co-linear is because you made them so. Any two points $(a,b_1)$ and $(a, b_2)$ will be along the line x=a no matter what the y-values are.

Let me ask you this. What does f(-.44) represent? What about f'(-.44)? Finally what does f''(-.44) represent?
 

DeusAbscondus

Active member
Jun 30, 2012
176
Sorry to ask for more explaining but I'm just not getting something. What features do you mean here?

It this example you posted f(x), f'(x) and f'''(x) as graphs. You solved for f'(x)=0 which is at x = -.44. Then you found f(-.44) and f''(-.44) and plotted those points, and finally connected all three points with a dotted line.

Again, the reason they are co-linear is because you made them so. Any two points $(a,b_1)$ and $(a, b_2)$ will be along the line x=a no matter what the y-values are.

Let me ask you this. What does f(-.44) represent? What about f'(-.44)? Finally what does f''(-.44) represent?
No need for apologies! Au contraire it is I, if anyone should be apologizing, for lack of clarity.

I'm going to give more thought to the framing of this question; it will be a good exercise.
As a result of the thinking, perhaps the question will answer itself; or perhaps the thinking will reveal to me that I am missing a more fundamental property of "true by definition" (in fact, i'm beginning to think this is precisely what I am overlooking, and thereby making mystery where there is none: like a kitten playing with a ball of string!)

Thanks for bearing with my faltering steps and gaps in logic.
Deus Abs

PS incidentally, I was nearly struck by an Eastern Brown snake today, the second-deadliest land snake on the Blue Planet. I rescue Australian fauna as a volunteer and this guy, a 1.7 meter exemplar of the species, had installed himself a bit too close to a member of public's front verandah. When I tried to move him on, he wasn't happy about relocation plans not of his own devising and struck like lightening. My reflexes are very good, but I still just managed to withdraw my leg in time to avoid a bite, but not fast enough to avoid letting him make contact with my be-denimed leg. I finally managed to bag him and then release him 500 metres down the road, in a vacant land corridor.
Close call; gives life a salty taste.
 
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Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
What inference can be drawn for all \(f(x)\) when \(f(x)\) is twice-differentiable and
\(f(x)\) contains at least one inflection point........
Hi DeusAbscondus, :)

As Jameson had pointed out previously, the three points,

1) Inflection point of \(f(x)\)

2) Critical point of \(f'(x)\)

3) The root of \(f''(x)\)

are collinear since all have the same \(x\) coordinate.

If you are looking for some general formulas for the \(y\) coordinates of the inflection point of \(f(x)\) and the critical point of \(f'(x)\) then the following calculations may help. :)

Take a general cubic polynomial,

\[f(x)=ax^3+bx^2+cx+d\]

When, \(f''(x)=0\Rightarrow x=-\dfrac{b}{3a}\).

\[\therefore f'\left(-\frac{b}{3a}\right)=c-\frac{{b}^{2}}{3a} \mbox{ and }f\left(-\frac{b}{3a}\right)=d-\frac{b\,c}{3\,a}+\frac{2\,{b}^{3}}{27\,{a}^{2}}\]

These are the \(y\) coordinates of the critical point of \(f'(x)\) and the inflection point of \(f(x)\) (for a general cubic polynomial) respectively.

Kind Regards,
Sudharaka.
 
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DeusAbscondus

Active member
Jun 30, 2012
176
Hi DeusAbscondus, :)

As Jameson had pointed out previously, the three points,

1) Inflection point of \(f(x)\)

2) Critical point of \(f'(x)\)

3) The root of \(f''(x)\)

are collinear since all have the same \(x\) coordinate.

If you are looking for some general formulas for the \(y\) coordinates of the inflection point of \(f(x)\) and the critical point of \(f'(x)\) then the following calculations may help. :)

Take a general cubic polynomial,

\[f(x)=ax^3+bx^2+cx+d\]

When, \(f''(x)=0\Rightarrow x=\dfrac{b}{3a}\).

\[\therefore f'\left(\frac{b}{3a}\right)=c+\frac{{b}^{2}}{a} \mbox{ and }f\left(\frac{b}{3a}\right)=d+\frac{b\,c}{3\,a}+ \frac{4\,{b}^{3}}{27\,{a}^{2}}\]

These are the \(y\) coordinates of the critical point of \(f'(x)\) and the inflection point of \(f(x)\) (for a general cubic polynomial) respectively.

Kind Regards,
Sudharaka.
Sudharaka! this is splendid: thank you kindly!

In all fairness to Jameson, I had not even come close to formulating my question coherently and he battled manfully, with one hand tied behind his back, trying to help me; but here you have divined the *kind* of thing I was fumbling for: some formal statement (beyond a mere restatement of the definition of "inflection point", "root" and "min. turning point") encompassing the relationship between $$f^{n}(x)$$
$$f^{n+1}(x)\text{,and}$$
$$f^{n+2}(x)$$

And I think for now, I will consider myself enriched perhaps beyond my present means to enjoy my wealth, and return to the bread-and-water diet of problem solving adapted to my level.(Whew)

But, just in parting, I could not *not* ask the question, as it was bugging me: I want to have a sense of what all these strange objects and quanities are; so much so that I want to combine my maths journey with physics, so as to have some vectors or other physical phenomena to apply these formulae to (in fact, I *am* going to extend my formal studies to beginners' physics next year, so inspired am I by my introduction to Calculus)

thanks again Sudharaka, and Jameson.
DeusAbs
 

DeusAbscondus

Active member
Jun 30, 2012
176
Hi DeusAbscondus, :)

As Jameson had pointed out previously, the three points,

1) Inflection point of \(f(x)\)

2) Critical point of \(f'(x)\)

3) The root of \(f''(x)\)

are collinear since all have the same \(x\) coordinate.

If you are looking for some general formulas for the \(y\) coordinates of the inflection point of \(f(x)\) and the critical point of \(f'(x)\) then the following calculations may help. :)

Take a general cubic polynomial,

\[f(x)=ax^3+bx^2+cx+d\]

When, \(f''(x)=0\Rightarrow x=-\dfrac{b}{3a}\).

\[\therefore f'\left(-\frac{b}{3a}\right)=c-\frac{{b}^{2}}{3a} \mbox{ and }f\left(-\frac{b}{3a}\right)=d-\frac{b\,c}{3\,a}+\frac{2\,{b}^{3}}{27\,{a}^{2}}\]

These are the \(y\) coordinates of the critical point of \(f'(x)\) and the inflection point of \(f(x)\) (for a general cubic polynomial) respectively.

Kind Regards,
Sudharaka.
It can't escape notice on even a casual glance at these equations (for the first time in my case) the ultimate kinship of derivation between the above and that which we use to find the vertex of a parabola: could you make a comment or two touching on this "common derivation" for want of better words.

thanks again for your kind help,

Deus Abs
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
It can't escape notice on even a casual glance at these equations (for the first time in my case) the ultimate kinship of derivation between the above and that which we use to find the vertex of a parabola: could you make a comment or two touching on this "common derivation" for want of better words.
If $f(x)=ax^2+bx+c$ and $(x_0,y_0)$ are the coordinates of the vertex, then $f'(x_0)=2ax_0+b=0$, from where $x_0=-b/(2a)$.
 

DeusAbscondus

Active member
Jun 30, 2012
176
If $f(x)=ax^2+bx+c$ and $(x_0,y_0)$ are the coordinates of the vertex, then $f'(x_0)=2ax_0+b=0$, from where $x_0=-b/(2a)$.
Many thanks Evgeny,

Could you (or anyone browsing) explain to me this notation, which is totally unfamiliar to me:
$$(x_0, y_0)$$
I suspect it means that it means $$x\ne0, y\ne0$$ since, otherwise, if $x=0$ this would make $a=0$ in the above equations, which you can't have in a quadratic or cubic, (right?). But excluding $x=0$ would prohibit the occurrence of vertices which are at $(0,0)$, which is an obvious nonsense, so I'm unsure.

*scratching head*(Worried)

Deus Abs
(stumbling through an eerily-half-lit demi-monde landscape, filled with strange new geographical features like curves which are continuous everywhere but no-where differentiable, momentarily catching glimpses through parting mists of snow-capped, impersonal peaks of exquisite beauty with ugly names like: Mount $f^{n+2}(x)$ :cool: )
 
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Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi DeusAbscondus. :)

Many thanks Evgeny,

Could you (or anyone browsing) explain to me this notation, which is totally unfamiliar to me:
$$(x_0, y_0)$$
I suspect it means that it means $$x\ne0, y\ne0$$ since, otherwise, if $x=0$ this would make $a=0$, which you can't have in a quadratic or cubic, (right?). But excluding $x=0$ would prohibit the occurrence of vertices which are at $(0,0)$, which is an obvious nonsense, so I'm unsure.

*scratching head*(Worried)
As Evgeny.Makarov had explained \((x_0, y_0)\) represents the coordinates of the vertex of the quadratic curve. You have already used this notation in your post #6;

DeusAbscondus said:
\(C=(0.33,3.67)\) being Minimum Turning Point on \(f'(x)\)
What Evgeny.Makarov had done is consider a general quadratic curve; \(f(x)=ax^2+bx+c\) and taken it's vertex(turning point which can be a minimum or a maximum) as \((x_0,y_0)\).

Deus Abs
(stumbling through an eerily-half-lit demi-monde landscape, filled with strange new geographical features like curves which are continuous everywhere but no-where differentiable, momentarily catching glimpses through parting mists of snow-capped, impersonal peaks of exquisite beauty with ugly names like: Mount $f^{n+2}(x)$ :cool: )
Talking about functions which are "continuous everywhere but no-where differentiable" you might be interested in the Weierstrass function.

Kind Regards,
Sudharaka.
 

DeusAbscondus

Active member
Jun 30, 2012
176
[/QUOTE]
What Evgeny.Makarov had done is consider a general quadratic curve; \(f(x)=ax^2+bx+c\) and taken it's vertex(turning point which can be a minimum or a maximum) as \((x_0,y_0)\).



Talking about functions which are "continuous everywhere but no-where differentiable" you might be interested in the Weierstrass function.

Kind Regards,
Sudharaka.[/QUOTE]

Thanks kindly Sudharaka,
As for the Weierstrass function, it was reading about this function where I got the terminology in the first place. Very strange beast: I wonder what they made of it in the 1870s, a century before the breakthrough represented by chaos theory and the science of fractals?

As for the notation used by Evgeny: I am *still* unsure as to the precise value of the sub-scripted $0$ in $$(x_0, y_0)$$.
Are you saying that the zero simply means that the co-ordinate or abscissa preceding it is a value of the vertex to a quadratic curve?

If so, that makes perfect sense.
(I just get nervous around new $0s$ popping up in unfamiliar places :d)

DeusAbs
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
DeusAbscondus said:
Thanks kindly Sudharaka,
As for the Weierstrass function, it was reading about this function where I got the terminology in the first place. Very strange beast: I wonder what they made of it in the 1870s, a century before the breakthrough represented by chaos theory and the science of fractals?

As for the notation used by Evgeny: I am *still* unsure as to the precise value of the sub-scripted $0$ in $$(x_0, y_0)$$.
Are you saying that the zero simply means that the co-ordinate or abscissa preceding it is a value of the vertex to a quadratic curve?

If so, that makes perfect sense.
(I just get nervous around new $0s$ popping up in unfamiliar places :d)

DeusAbs
The subscript \(0\) in \((x_0,y_0)\) signifies nothing special. If you are uneasy about this notation you may well use, \((a,b)\) to represent the vertex point. But when you want to specify a large number of different points the sub-script notation comes in handy. For example you have twenty points that you want to label on the curve. Then you will run out of letters if you try to mark them as, \((a,b),\,(c,d)\) and so on. Instead you can mark them,

\[(x_0,y_0),\,(x_1,y_1),\,(x_2,y_2),\,(x_3,y_3),\, \cdots,\,(x_{20},y_{20})\]
 

DeusAbscondus

Active member
Jun 30, 2012
176
Thanks again.
I think part of challenge for someone like me communicating with others on a site like this (and which was highlighted recently by a comment made by Captain Black, when asking me to be specific about 'what i already know') is the patchy nature of my mathematical maturation process: I have bypassed significant swathes of what would be considered normal developmental steps in that maturation process, because of the utterly eccentric and personal way that I have acquired the little maths that I *do* have.

So, to bring it back to this instance: you guys probably assume (and naturally enough!) that a guy like me asking questions on a calculus forum, who apparently shows growth and awareness and skills in certain aspects, has already met such a humble, work-a-day piece of notation as the one you describe above, but i have not!

Indeed, that is not entirely true: while wading through Penrose's Emporer's New Mind I constantly came across notation like that, but, to my eternal frustration I could never work out what it meant (along with heaps of other stuff that was entirely opaque to me)

Anyway, I continue, a squillionth of a unit more skilled and knowledgeable every day!
Thanks to you guys (and my dogged persistence)

Deus Abs
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
The subscript \(0\) in \((x_0,y_0)\) signifies nothing special. If you are uneasy about this notation you may well use, \((a,b)\) to represent the vertex point. But when you want to specify a large number of different points the sub-script notation comes in handy. For example you have twenty points that you want to label on the curve. Then you will run out of letters if you try to mark them as, \((a,b),\,(c,d)\) and so on. Instead you can mark them,

\[(x_0,y_0),\,(x_1,y_1),\,(x_2,y_2),\,(x_3,y_3),\, \cdots,\,(x_{20},y_{20})\]
Yes. In this case, the subscript 0 is not an operation; rather, it is a part of the variable name. For example, in many programming languages variable names may consist of letters, digits and underscores.

I wanted to distinguish the variable x used in the definition of f(x), which can range over all real numbers, from the abscissa of the parabola's vertex, which is a constant. I often encountered a situation when an author wanted to consider a specific value of a variable and he/she gave it a new name by appending a subscript to the variable's name. So this is largely a matter of style.
 

DeusAbscondus

Active member
Jun 30, 2012
176
Yes. In this case, the subscript 0 is not an operation; rather, it is a part of the variable name. For example, in many programming languages variable names may consist of letters, digits and underscores.

I wanted to distinguish the variable x used in the definition of f(x), which can range over all real numbers, from the abscissa of the parabola's vertex, which is a constant. I often encountered a situation when an author wanted to consider a specific value of a variable and he/she gave it a new name by appending a subscript to the variable's name. So this is largely a matter of style.
Thank you Evgeny.
I've made a Memory Card with these facts written up.

Deus Abs
 
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