Gap in understanding Taylor series

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roineust

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I've been studying in Khan academy: AP®︎/College Calculus AB and then also AP®︎/College Calculus BC.

Up until the sub-unit I'm in right now, which is: Finding Taylor polynomial approximations of functions, which is inside Unit 10: Infinite sequences and series, I've felt that everything taught to me all the way back to the beginning, which for me, took many months of studying, all material was taught from a fundamental perspective. There was nothing that was just laid out as facts to memorize, without explaining the full process that motivationally build up to a mathematical concept.

Suddenly i watch the Taylor series first videos and i think that I've unintentionally skipped several videos before it, but presumably no.

Yes, i feel that i understood all the material in Unit 10: Infinite sequences and series, up to that point, but no, i'm far from fully understanding what is thrown at me here.

What is this? Why do this expansion? How this has developed into this polynomial structure? What are the most basic examples that brought this concept into development? When is this helpful?

Even when going into a Google and YouTube and Open AI to search, i couldn't find an explanation that felt to me is a ground up explanation.

The only explanation i could find, which also seems to be only partially intuitive and partially ground up is:

medium.com

Taylor series- for dummies, by a dummy

What is Taylor series? Why is approximation important?
medium.com medium.com

Taylor series helps when trying to solve more complex functions? Like which functions? Like Sin or Exponent functions? Then why is it that the polynomial that approximates these functions, is built from the exact same functions that make the complexity problem in the first place? In which way are these functions complex to solve? In what way do they present a complexity problem that raises or raised the need to develop the Taylor polynomial structure? Why is the saying that this is the way computers calculate some functions come into context?

And how was that polynomial structure discovered and developed in the first place? Why does this polynomial structure approximate other functions? How was the mechanics of this structure developed? What's the whole thing with the factorial and why does the word normalization come into context?

How come i go through a course for many months and everything seems more or less explained to me at an intuitive level and then suddenly at once, there is such a gap in the level and depth of explanation?

Have i reached some transparent mysterious gate in the studies of mathematics, where only people with higher IQ than me are allowed to continue through?
 
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pka said:
Please state a carefully formed question without all the blather.
Click to expand...
Very well, i will ask the question without regarding the prattle in your response and you better be careful with the way you read, since it is a quotation from the first post:

" Taylor series helps when trying to solve more complex functions? Like which functions? Like Sin or Exponent functions? Then why is it that the polynomial that approximates these functions, is built from the exact same functions that make the complexity problem in the first place? "

I'm not a young person, please skip the parts of enunciation that are fit for younger people.
 
roineust said:
when trying to solve more complex functions
Click to expand...
What do you mean by "solve... functions"?!

I know how to solve equations.

Solving functions would be a new endeavor for me !

I could derive the functional values (approximately) around a given point - if the Taylor series expansion given to me. My grandson calls this invoking Taylor Swift (possibly because all these happen swiftly.

In general, Taylor series expansion of a function [f(x)] will "unpack" the function (around a given point xo

in terms of the value of the first derivative of the function at xo and​
​
,in terms of the value of the second derivative of the function at xo and​
​
in terms of the value of the third derivative of the function at xo and ..... so on....​

Thus the expansion provides an approximation - most of the time polynomial approximation - of the function around a given point. This is used extensively in engineering.
 
khansaheb said:
Please read:

How do you find the Taylor series for ln(x) about the value x=1? | Socratic

firstly we look at the formula for the Taylor series, which is: f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n which equals: f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/(2!) + (f'''(a)(x-a)^3)/(3!) + ... So you would like to solve for f(x) = ln(x) at x=1 which I assume mean centered at 1 of which you...
socratic.org socratic.org
Click to expand...
You don't seem to understand that i don't understand why this whole process is done.

You want to solve the equation ln(1), by creating some polynomial series structure that includes within it the equation ln(1)? So what does this helps to solve ln(1), if you have in the solution ln(1), which you want to solve in the first place?
 
roineust said:
You don't seem to understand that i don't understand why this whole process is done.

You want to solve the equation ln(1), by creating some polynomial series structure that includes within it the equation ln(1)? So what does this helps to solve ln(1), if you have in the solution ln(1), which you want to solve in the first place?
Click to expand...

We already know what ln(1) is; it's 0, from the definition of the ln function.

But suppose you want to find ln(1.3). How would you find that?

I suppose you'd just pick up your calculator and type it in, right?

The trick is: Your calculator may well be using this Taylor series to do its calculation! (There are additional tricks it will do to be more efficient.)

That's why Taylor series are done (and for many more advanced problems). It allows you to calculate things (approximately) that otherwise couldn't be calculated, by reducing it to a polynomial.

To be more specific, [math]\ln(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{(x-1)^n}{n}=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-\dots[/math]
So [math]\ln(1.3)=(1.3-1)-\frac{1}{2}(1.3-1)^2+\frac{1}{3}(1.3-1)^3-\frac{1}{4}(1.3-1)^4+\frac{1}{5}(1.3-1)^5\dots\\=0.3-\frac{1}{2}0.3^2+\frac{1}{3}0.3^3-\frac{1}{4}0.3^4+\frac{1}{5}0.3^5\dots\\=0.3-0.045+0.009-0.002025+0.000486-\dots\approx0.262461[/math]
The actual value, according to my calculator, is 0.262364... . If we take more terms, we'll be more accurate; and there are ways to find how accurate you are.

Does that help?
 
roineust said:
You don't seem to understand that i don't understand why this whole process is done.

You want to solve the equation ln(1), by creating some polynomial series structure that includes within it the equation ln(1)? So what does this helps to solve ln(1), if you have in the solution ln(1), which you want to solve in the first place?
Click to expand...
No ..... you want to calculate the value of ln(1.01) knowing the value of ln(1) and d/dx [ln(x)@(x=1)] and d2/dx2[ln(x)@(x=1)]
and so on....
 
roineust said:
You don't seem to understand that i don't understand why this whole process is done.
You want to solve the equation ln(1), by creating some polynomial series structure that includes within it the equation ln(1)? So what does this helps to solve ln(1), if you have in the solution ln(1), which you want to solve in the first place?
Click to expand...
Now THAT is a well and simply-formed question!
I began teaching in the mid-1960's. We would tell students that [imath]\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}[/imath]
is rationalized because it is a deal easier to divide a radical by a whole number than the other way around.
But mind you that was ten years before calculators were available at a reasonable cost.
The use of a Taylor's series is somewhat analogous. At one time we used Taylor series to approximate values of functions at points.

[imath][/imath][imath][/imath][imath][/imath]
 
Dr.Peterson said:
We already know what ln(1) is; it's 0, from the definition of the ln function.

But suppose you want to find ln(1.3). How would you find that?

I suppose you'd just pick up your calculator and type it in, right?

The trick is: Your calculator may well be using this Taylor series to do its calculation! (There are additional tricks it will do to be more efficient.)

That's why Taylor series are done (and for many more advanced problems). It allows you to calculate things (approximately) that otherwise couldn't be calculated, by reducing it to a polynomial.

To be more specific, [math]\ln(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{(x-1)^n}{n}=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-\dots[/math]
So [math]\ln(1.3)=(1.3-1)-\frac{1}{2}(1.3-1)^2+\frac{1}{3}(1.3-1)^3-\frac{1}{4}(1.3-1)^4+\frac{1}{5}(1.3-1)^5\dots\\=0.3-\frac{1}{2}0.3^2+\frac{1}{3}0.3^3-\frac{1}{4}0.3^4+\frac{1}{5}0.3^5\dots\\=0.3-0.045+0.009-0.002025+0.000486-\dots\approx0.262461[/math]
The actual value, according to my calculator, is 0.262364... . If we take more terms, we'll be more accurate; and there are ways to find how accurate you are.

Does that help?
Click to expand...
Yes, this helps somewhat not wholly, since i still don't understand every bit of how the Taylor series was discovered from the very beginning. I was under the impression that i understood every bit, when it was the idea of the limit, the idea of the integral and also the series structure, up to this point. In a way, the mathematical historical perspective is suddenly thrown away from the window. History in the sense of mathematics, not in the sense of history.

In my mind, this is a very clear situation, in which after many months of studying the Khan Academy AB and BC calculus courses, and feeling that i understand not only the 'how' of mathematical concepts and structures presented to me, but also the 'why', suddenly, not even at a gradual creeping pace, but at once, from one moment to the next, a structure is thrown at me, the Taylor series, that while i can maybe still repeat the 'how' about that structure, the' how' being the 'algorithm', the 'cooking recipe', the 'mechanics', the reason for this 'how' to look and work the way it does, in other words, the 'why', has been taken away from me, and taken away as a method of teaching, for the first time in a whole long course. Is it supposed to be this way for calculus math students when they reach the Taylor series?
Dr.Peterson said:
We already know what ln(1) is; it's 0, from the definition of the ln function.

But suppose you want to find ln(1.3). How would you find that?

I suppose you'd just pick up your calculator and type it in, right?

The trick is: Your calculator may well be using this Taylor series to do its calculation! (There are additional tricks it will do to be more efficient.)

That's why Taylor series are done (and for many more advanced problems). It allows you to calculate things (approximately) that otherwise couldn't be calculated, by reducing it to a polynomial.

To be more specific, [math]\ln(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{(x-1)^n}{n}=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-\dots[/math]
So [math]\ln(1.3)=(1.3-1)-\frac{1}{2}(1.3-1)^2+\frac{1}{3}(1.3-1)^3-\frac{1}{4}(1.3-1)^4+\frac{1}{5}(1.3-1)^5\dots\\=0.3-\frac{1}{2}0.3^2+\frac{1}{3}0.3^3-\frac{1}{4}0.3^4+\frac{1}{5}0.3^5\dots\\=0.3-0.045+0.009-0.002025+0.000486-\dots\approx0.262461[/math]
The actual value, according to my calculator, is 0.262364... . If we take more terms, we'll be more accurate; and there are ways to find how accurate you are.

Does that help?
Click to expand...
 
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roineust said:
Yes, this helps somewhat not wholly, since i still don't understand every bit of how the Taylor series was discovered from the very beginning.
Click to expand...
I'm not sure that a complete history is the same as the "why" you asked about. (The word "why" has several meanings, so perhaps you need to clarify which you were asking for.)

In any case, I don't expect to fully understand anything when first introduced; there's always something deeper that is left out, for good reasons.
roineust said:
I was under the impression that i understood every bit, when it was the idea of the limit, the idea of the integral and also the series structure, up to this point.
Click to expand...
I don't know that I do.
roineust said:
suddenly, not even at a gradual creeping pace, but at once, from one moment to the next, a structure is thrown at me, the Taylor series, that while i can maybe still repeat the 'how' about that structure, the' how' being the 'algorithm', the 'cooking recipe', the 'mechanics', the reason for this 'how' to look and work the way it does, in other words, the 'why', has been taken away from me, and taken away as a method of teaching, for the first time in a whole long course.
Click to expand...

This would seem to be a criticism of a particular course by a particular teacher. Why are you not complaining to the author(s)? Why are you talking as if everyone fails to explain why, and/or in what historical context, Taylor series came about? I'm surprised that anyone would introduce this topic with no motivation at all, which seems to be what you are saying.

Anyway, what are you still lacking?
 
roineust said:
In my mind, this is a very clear situation, in which after many months of studying the Khan Academy AB and BC calculus courses, and feeling that i understand not only the 'how' of mathematical concepts and structures presented to me, but also the 'why', suddenly, not even at a gradual creeping pace, but at once, from one moment to the next, a structure is thrown at me, the Taylor series, that while i can maybe still repeat the 'how' about that structure, the' how' being the 'algorithm', the 'cooking recipe', the 'mechanics', the reason for this 'how' to look and work the way it does, in other words, the 'why', has been taken away from me, and taken away as a method of teaching, for the first time in a whole long course. Is it supposed to be this way for calculus math students when they reach the Taylor series?
Click to expand...
I will preface this by saying that I am no fan of the Kahn Academy. I have seen students with the same complaints as you have.
Here is a textbook the the 1973 edition of which I consider the best calculus text available. Buy from the used book market for less than $20 US.
The author was the president of the MAA and finished his career at UT Austin. In it the discussion of Taylor series is one of the very last topics given.
 
roineust said:
Yes, this helps somewhat not wholly, since i still don't understand every bit of how the Taylor series was discovered from the very beginning. I was under the impression that i understood every bit, when it was the idea of the limit, the idea of the integral and also the series structure, up to this point. In a way, the mathematical historical perspective is suddenly thrown away from the window. History in the sense of mathematics, not in the sense of history.
Click to expand...
Here is what Wikipedia has to say about the historical background:

"The first approach to implicit usage of the concept of a function in tabular form (shadow length depending on the time of day, chord lengths depending on the central angle, etc.) can already be seen in antiquity. The first evidence of an explicit definition of a function can be found in Nicholas of Oresme (14th century AC), who graphically represented the dependencies of changing quantities (heat, movement, etc.) using perpendicular lines. At the beginning of the process of developing a rigorous understanding of functions were Descartes and Fermat, who developed the analytical method of introducing functions with the help of the variables recently introduced by Vieta. Functional dependencies should therefore be represented by equations such as [imath] y=x^{2} [/imath]. This naive concept of functions was retained in school mathematics until well into the second half of the 20th century. The first description of functions according to this idea comes from Gregory in his book Vera circuli et hyperbolae quadratura, published in 1667. The technical term 'function' appears for the first time in 1673 in a manuscript by Leibniz, who also used the terms 'constant', 'variable', 'ordinate', and 'abscissa' in his treatise De linea ex lineis numero infinitis ordinatim ductis, 1692. The concept of functions is detached from geometry and transferred to algebra in the correspondence between Leibniz and Johann Bernoulli. Bernoulli presents this development in several contributions at the beginning of the 18th century. Leonhard Euler, a student of Johann Bernoulli, further specified functions in his book Introductio in analysin infinitorum in 1748. We can find two different explanations for a function in Euler: On the one hand, every 'analytic expression' in x represents a function, on the other hand, y(x) is defined in the coordinate system by a freehand drawn curve. In 1755 he reformulated these ideas without using the term 'analytic expression'. He also coined the notation f(x) as early as 1734. Euler distinguishes between unique and ambiguous functions. The inverse of the normal parabola, in which every non-negative real number is assigned both its positive and its negative root, is also permitted as a function according to Euler. For Lagrange, only functions that are defined by power series are permissible, as he states in his Théorie des fonctions analytiques in 1797."
roineust said:
In my mind, this is a very clear situation, in which after many months of studying the Khan Academy AB and BC calculus courses, and feeling that i understand not only the 'how' of mathematical concepts and structures presented to me, but also the 'why', suddenly, not even at a gradual creeping pace, but at once, from one moment to the next, a structure is thrown at me, the Taylor series, that while i can maybe still repeat the 'how' about that structure, the' how' being the 'algorithm', the 'cooking recipe', the 'mechanics', the reason for this 'how' to look and work the way it does, in other words, the 'why', has been taken away from me, and taken away as a method of teaching, for the first time in a whole long course. Is it supposed to be this way for calculus math students when they reach the Taylor series?
Click to expand...
Lagrange uses power series for the first time to define functions,
[math]f(x)=\sum_{n=0}^\infty a_n(x-p)^n .[/math]So in a sense, it all started with power series.
pka said:
I will preface this by saying that I am no fan of the Kahn Academy. I have seen students with the same complaints as you have.
Here is a textbook the the 1973 edition of which I consider the best calculus text available. Buy from the used book market for less than $20 US.
The author was the president of the MAA and finished his career at UT Austin. In it the discussion of Taylor series is one of the very last topics given.
Click to expand...

Technically, we get the Taylor series e.g. by the fundamental theorem of calculus, and continued integration by parts.

(I'm new and have difficulties using LaTeX here, esp. the align* environment, so I couldn't insert the formulas. And I am not sure whether I am allowed to link to a text I wrote on another website.)
 
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