Taylor polynomials doubt

[joantica_]

I interpret their "seven" as being the "n" in...

[math]{T_n}\left( x \right) = \sum\limits_{i = 0}^n {\frac{{{f^{\left( i \right)}}\left( a \right)}}{{i!}}{{\left( {x - a} \right)}^i}}[/math]
...otherwise the question just doesn't seem to make sense (to me anyway)

Most pages I've just looked up talk about the n^th degree Taylor polynomial in general terms without specifying that the n^th derivative must be non-zero. For example http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx @tkhunny please can you cite a reference to a definition that states there must be a non-zero coefficient to the last, n^th, term?

Apart from the discussion of degrees, could you confirm me what I posted, that according to you, then p(x) is an example of a function which has p(x) as its Taylor polinomial of degree seven - regardless the value of a (in this case a = 4) -. Yes?, if I haven't interpreted your response improperly

 

[joantica_]

Anyway - if anyone is interested here's one way to get infinite different functions f(x) that have p(x) as their "n=7" Taylor series (carefully avoiding the word degree )

f(x) = p(x) + c*[ cos(x-4) - ( 1 - (x-4)^2/2! + (x-4)^4/4! - (x-4)^6/6! ) ]

The expression within square brackets [] has no Taylor series for the first 7 terms, so "c" can have any value. This only works around x=4 ( a=4 ).

This works because "( 1 - (x-4)^2/2! + (x-4)^4/4! - (x-4)^6/6! )" are the first few terms of the Taylor series for cos(x-4).

How do you prove that? Just considering the taylor series for cos(x-4) as you said?

 

[Cubist]

How do you prove that? Just considering the taylor series for cos(x-4) as you said?

I would not fancy doing the Taylor expansion of that by hand. So yes, I'd argue it with logic. I did check it in "maxima" (a computer algebra system) and it worked.

However, I recommend that you stick to using the simple example of p(x) in your answer. Post#20 is mostly for your interest/ info.

And to be on the safe side you could clearly state the assumption that the words "degree seven" are assumed to mean "n=7" in the Taylor series.

 

[tkhunny]

Where does it mention Taylor series?

Where did it happen, between Algebra I and Taylor Series, that we redefined the degree of a polynomial?

Where does the original problem statement mention "n=7"? It's fine if you want to define a different question and then answer the new question. I don't have a problem with that. It's just not an answer to the original question.

 

[Cubist]

Where did it happen, between Algebra I and Taylor Series, that we redefined the degree of a polynomial?

I personally don't claim to be an authority. And I definately respect your opinion. But isn't it possible that the word "degree", in this context, refers to "n" in the definition of the Taylor polynomial? ie before a certain "f" has been plugged in. At this stage it is a polynomial with an x^n term.

You'll probably agree that it's useful to have the knowledge of what value of "n" was used even if it generated a zero coefficient. (For calculating extra terms, or error bounds purposes.)

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I have looked up the first page Google results for both "taylor polynomial degree" and "degree of taylor polynomial" and they didn't contain a clear definition for the word "degree" in this context (or at least I didn't think so). However the following page, further down, implied the above meaning:-

http://www.math.smith.edu/~rhaas/m114-00/chp4taylor.pdf

"the second degree Taylor Polynomial for sin x near 0 is P2(x)=x. It is rather disappointing that this turns out to be no different from P1 for sin x."

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I do think it's a great shame that there doesn't seem to be a "go to" source for precise mathematical definitions and nomenclature. Then everyone can be on the same page when communicating.