Taylor polynomials doubt

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joantica_

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Hi everyone!

I’m student, currently I’m at the University as I’m studying a degree in Maths & Statistics.

I have been recently working on Taylor Series, as a way to calculate limits and other applications. Anyway, days ago I started doing many exercises but there is one that I can’t resolve, because it includes a resolution method (I think) I have not directly worked about and everything I have tried didn’t work. The question is:
Consider the polynomial p(x) = 3 + 4x^2 - 5x^3. Can it be the Taylor polynomial of degree seven of a function in the point a=4?

Taylor polynomial:
f(a) + f’(a)*(x-a) + (f’’(a)/2!)*(x-a)^2 + ...

Any help will be well received, thank you!
 
How many derivatives has p(x)?

Are we SURE we're asking the right question? What's the point of making a Taylor Polynomial out of something that is already a polynomial?
 
Yes, is the right question, maybe it is erroneous (surely not) but it’s the question I have.
We’re given the polynomial p(x) and the question is exactly if exists any function with p(x) as its Taylor polynomial of seventh degree (and if so, give an example).
Derivatives of p(x):
p’(x) = 8x - 15x^2
p’’(x) = 8 - 30x
p’’’(x) = -30
p’’’’(x) = 0
...
...
 
joantica_ said:
Derivatives of p(x):
p’(x) = 8x - 15x^2
p’’(x) = 8 - 30x
p’’’(x) = -30
p’’’’(x) = 0
Click to expand...

OK, using the derivatives above, what would be the highest power of x in a 7th degree Taylor series of p(x) ?

In fact, you might want to go ahead and calculate the 7th degree Taylor series of p(x). I think this question might be aimed at showing you something about the Taylor series of a polynomial.
 
Notice that the question does NOT ask you to find the polynomial itself. It just asks "IF this third degree polynomial can be the Taylor polynomial of a seventh degree polynomial about x= 4".

The answer should be an immediate "NO"! The "Taylor polynomial of a polynomial, of degree n, about any "\(\displaystyle x= x_0\)", is a polynomial of degree n. In fact, you can find the "Taylor polynomial about x= 4" for this polynomial, not by differentiating, but by a simple change of variable. Let y= x- 4 so that x= y+ 4 and \(\displaystyle 3 + 4x^2 - 5x^3= 3+ 4(y+ 4)^2- 5(y+ 4)^3= 3+ 4y^2+ 32y+ 64+ 5y^3+ 60y^2+ 240y+ 320= 5y^3+ 64y^2+ 72y+ 387= 5(x- 4)^3+ 64(x- 4)^2+ 72(x- 4)+ 387\) (check arithmetic).
 
Note that post#1 didn't contain the whole question...

joantica_ said:
...if exists any function with p(x) as its Taylor polynomial of seventh degree (and if so, give an example)
Click to expand...



HallsofIvy said:
The answer should be an immediate "NO"! The "Taylor polynomial of a polynomial, of degree n, about any "\(\displaystyle x= x_0\)", is a polynomial of degree n. In fact, you can find the "Taylor polynomial about x= 4" for this polynomial, not by differentiating, but by a simple change of variable. Let y= x- 4 so that x= y+ 4 and \(\displaystyle 3 + 4x^2 - 5x^3= 3+ 4(y+ 4)^2- 5(y+ 4)^3= 3+ 4y^2+ 32y+ 64+ 5y^3+ 60y^2+ 240y+ 320= 5y^3+ 64y^2+ 72y+ 387= 5(x- 4)^3+ 64(x- 4)^2+ 72(x- 4)+ 387\) (check arithmetic).
Click to expand...

I get something different for the Taylor series of this polynomial. (The answer might be a surprise, and I was trying not to give it away to OP). I don't think that a Taylor series can be calculated by a shift of variable?
 
HallsofIvy said:
Notice that the question does NOT ask you to find the polynomial itself. It just asks "IF this third degree polynomial can be the Taylor polynomial of a seventh degree polynomial about x= 4".

The answer should be an immediate "NO"! The "Taylor polynomial of a polynomial, of degree n, about any "\(\displaystyle x= x_0\)", is a polynomial of degree n. In fact, you can find the "Taylor polynomial about x= 4" for this polynomial, not by differentiating, but by a simple change of variable. Let y= x- 4 so that x= y+ 4 and \(\displaystyle 3 + 4x^2 - 5x^3= 3+ 4(y+ 4)^2- 5(y+ 4)^3= 3+ 4y^2+ 32y+ 64+ 5y^3+ 60y^2+ 240y+ 320= 5y^3+ 64y^2+ 72y+ 387= 5(x- 4)^3+ 64(x- 4)^2+ 72(x- 4)+ 387\) (check arithmetic).
Click to expand...
Thank you for your response. However, I need to give an example if the function exists. I’m sorry I forgot to write it in the first message.
So, according to your message, you definitely conclude that it doesn’t exist.
I’m a bit frustrated because this type of exercises with “if so, give an example”, normally have solutions. Nevertheless, I know that it isn’t a rigid rule...
 
Cubist said:
Note that post#1 didn't contain the whole question...







I get something different for the Taylor series of this polynomial. (The answer might be a surprise, and I was trying not to give it away to OP). I don't think that a Taylor series can be calculated by a shift of variable?
Click to expand...
Thanks for your response, can I ask you What did you get? Have you demonstrated the existence of a function which has p(x) as its Taylor polynomial of seventh degree?
 
joantica_ said:
Thanks for your response, can I ask you What did you get? Have you demonstrated the existence of a function which has p(x) as its Taylor polynomial of seventh degree?
Click to expand...

Hi joantica, I could just tell you what I got but then you'd be less likely to remember what happens with the Taylor series of a polynomial in the future.

Is there a particular problem you're having with calculating this Taylor series, up to 7 terms at x=4 of 3 + 4x^2 - 5x^3 ? You've done some of the work already in post 3. I guess it can be a bit tedious expanding (x-4)^3, but it's worth the effort.
 


Cubist said:
Hi joantica, I could just tell you what I got but then you'd be less likely to remember what happens with the Taylor series of a polynomial in the future.

Is there a particular problem you're having with calculating this Taylor series, up to 7 terms at x=4 of 3 + 4x^2 - 5x^3 ? You've done some of the work already in post 3. I guess it can be a bit tedious expanding (x-4)^3, but it's worth the effort.
Click to expand...
Ahhh, I didn't know you needed that. Of course there is no problem calculating the Taylor series of p(x). At x = 4 I can only reach its third-degree polynomial:
-1774 + 240x + 4x^2 - 5x^3

From the fourth term the results are indefinitely equal to 0 as its derivative is 0. So, what's the reflexion about this at all?
 
joantica_ said:



Ahhh, I didn't know you needed that. Of course there is no problem calculating the Taylor series of p(x). At x = 4 I can only reach its third-degree polynomial:
-1774 + 240x + 4x^2 - 5x^3

From the fourth term the results are indefinitely equal to 0 as its derivative is 0. So, what's the reflexion about this at all?
Click to expand...

You must have made a little mistake somewhere. The answer is actually 3 + 4x^2 - 5x^3, in other words you get the same polynomial out - regardless of the value of a. NOTE This only works if you calculate more terms of the Taylor series than the degree of the polynomial.

So p(x) is itself an example of a function that gives p(x) as its Taylor polynomial !
 
Cubist said:
You must have made a little mistake somewhere. The answer is actually 3 + 4x^2 - 5x^3, in other words you get the same polynomial out - regardless of the value of a. NOTE This only works if you calculate more terms of the Taylor series than the degree of the polynomial.

So p(x) is itself an example of a function that gives p(x) as its Taylor polynomial !
Click to expand...
You're right, I made a mistake. So p(x) is itself an example of a function that gives p(x) as its Taylor polynomial, but not of degree seven (if what I'm saying is incorrect, tell me). The exercise asks me to find a function with p(x) as its Taylor polynomial of seventh degree.
One thing I have noticed is that calculating the Taylor series of k(x) = (x-4)^8 - 5x^3 + 4x^2 + 3, its Taylor series has p(x) as its third-degree polynomial and from the fourth derivative until the seventh, its coefficients are equal to 0, but at the eighth, the derivate isn't 0, having a polynomial of degree 8.
However, I don't know if considering k(x) we can say that its Taylor polynomial of degree third,fourth,...,seventh its p(x).

What do you think?
 
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joantica_ said:
The exercise asks me to find a function with p(x) as its Taylor polynomial of seventh degree.
Click to expand...

I think the degree seven in the question is referring to the number of terms to calculate in the Taylor series. This is different to the degree of the resultant polynomial p(x). At least that's my interpretation.


joantica_ said:
One thing I have noticed is that calculating the Taylor series of k(x) = (x-4)^8 - 5x^3 + 4x^2 + 3, its Taylor series has p(x) as its third-degree polynomial and from the fourth derivative until the seventh, its coefficients are equal to 0, but at the eighth, the derivate isn't 0, having a polynomial of degree 8.
However, I don't know if considering k(x) we can say that its Taylor polynomial of degree third,fourth,...,seventh its p(x).

What do you think?
Click to expand...

I like your thinking, but unfortunately I don't think that this idea will work. Having the "x^8" in there will mean that the fourth, fifth, etc derivatives will no longer be zero.

It would not surprise me if there are more functions that have p(x) as their 7th degree Taylor expansion. Maybe an infinite number! But I can't think of another one on the spur of the moment.
 
Cubist said:
I think the degree seven in the question is referring to the number of terms to calculate in the Taylor series. This is different to the degree of the resultant polynomial p(x). At least that's my interpretation.




I like your thinking, but unfortunately I don't think that this idea will work. Having the "x^8" in there will mean that the fourth, fifth, etc derivatives will no longer be zero.

It would not surprise me if there are more functions that have p(x) as their 7th degree Taylor expansion. Maybe an infinite number! But I can't think of another one on the spur of the moment.
Click to expand...

So, 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?
 
I do not believe the word "degree" ever would refer to a number of terms.

In my opinion, the ONLY thing to decide is what constitutes a TERM in a truncated Taylor Series.

If you have [math]p(x) = 1 + 7x + 2x^2 + 3x^3 + 0x^4 + 0x^5 + 0x^6 + 0x^7[/math], do you have a polynomial of degree 7 or not? Since Algebra I, we have not counted those terms with zero coefficients. Shall we start now?
 
tkhunny said:
I do not believe the word "degree" ever would refer to a number of terms.

In my opinion, the ONLY thing to decide is what constitutes a TERM in a truncated Taylor Series.

If you have [math]p(x) = 1 + 7x + 2x^2 + 3x^3 + 0x^4 + 0x^5 + 0x^6 + 0x^7[/math], do you have a polynomial of degree 7 or not? Since Algebra I, we have not counted those terms with zero coefficients. Shall we start now?
Click to expand...

Textually, the exercise asks: "Can p(x) be the Taylor polynomial of degree seven of a function at a = 4?" And considering "degree" as the degree of the polynomial p(x) it wouldn't make sense.
p(x) = 3 + 4x^2 - 5x^3
 
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 nth degree Taylor polynomial in general terms without specifying that the nth 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, nth, term?
 
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 :censored: :))

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).
 
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