Welcome to our community

Be a part of something great, join today!

Taylor Series

OhMyMarkov

Member
Mar 5, 2012
83
Hello Everyone!

Suppose $f(x)$ can be written as $f(x)=P_n(x)+R_n(x)$ where the first term on the RHS is the Taylor polynomial and the second term is the remainder.
If the sum $\sum _{n=0} ^{\infty} = c_n x^n$ converges for $|x|<R$, does this mean I can freely write $f(x)=\sum _{n=0} ^{\infty} = c_n x^n$?

Can I also use the fact that the sum of continuous functions over a domain (in this case, $|x|<R$) is continuous, and that the sum of differentiable functions over a domain is differentiable?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Could you rewrite your question? For example:

Hello Everyone! If the sum $\sum _{n=0} ^{\infty} = c_n x^n$
What does the left side mean? What is $c_n$?

Thanks.
 

OhMyMarkov

Member
Mar 5, 2012
83
$P_n(x)=\sum _{k=0} ^n c_k x^k$ is the Taylor polynomial. the $c_n$s are the coefficients.



EDIT: Ah, that was an honest typo, I meant the following:

Can I freely write $f(x)=\sum _{n=0} ^{\infty} c_n x^n$?
 

chisigma

Well-known member
Feb 13, 2012
1,704
... I meant the following...

... can I freely write $f(x)=\sum _{n=0} ^{\infty} c_n x^n$?
If the series converges inside and diverges outside a circle of radius R, then for $|x|<R$ the answer is YES!... an interesting question could be: and for $|x|=R$?... in this case the answer is: in the points on the circle where the series converges YES!...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
If the series converges inside and diverges outside a circle of radius R, then for $|x|<R$ the answer is YES!... an interesting question could be: and for $|x|=R$?... in this case the answer is: in the points on the circle where the series converges YES!...
How is important the concept is demonstrated by the following 'beatiful example'. In order to 'complete' [improbable!(Thinking)..] the discussion strated in...

http://www.mathhelpboards.com/f13/never-ending-dispute-2060/

... why don't try to asnswer to the question: what is $\varphi(x)=x^{x}$ for $x=0$?... avoiding wasting time answering to assertions like '$0^{0}$ is a nonsense' or something like that, we first use the identity $\displaystyle x^{x}= e^{x\ \ln x}$, so that we have to arrive at the value of the function $x\ \ln x$ for $x=0$. At first it seems that we are in the same situation because $0 * - \infty$ is 'indeterminate' but we don't decourage and search the Taylor expansion of the function $x\ \ln x$ around $x=1$. Without dreat effort we find...

$\displaystyle \psi (x)= x\ \ln x = (x-1) + \sum_{n=1}^{\infty} (-1)^{n}\ \frac{(x-1)^{n}}{n\ (n-1)}$ (1)

Very well!... it is relatively easy to 'discover' that the (1) converge inside a circle of radious R=1, but wht does it happen in $x=0$ that is on the circle?... in little time we obtain...

$\displaystyle \psi(0)= -1 + \frac{1}{2 \cdot 1} + \frac{1}{3 \cdot 2} + ... + \frac{1}{n\ (n-1)} + ...$ (2)

Now is...

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\ (n-1)} = \sum_{n=2}^{\infty} (\frac{1}{n-1}-\frac{1}{n}) = 1$ (3)


... so that is $\psi(0)=0 \implies \varphi(0)= 1$... one more dart on target! (Wink)...


Kind regards


$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Suppose $f(x)$ can be written as $f(x)=P_n(x)+R_n(x)$ where the first term on the RHS is the Taylor polynomial and the second term is the remainder.
If the sum $\sum _{n=0} ^{\infty} c_n x^n$ converges for $|x|<R$, does this mean I can freely write $f(x)=\sum _{n=0} ^{\infty} c_n x^n$?
In order to be able to write $f(x)=\sum _{n=0} ^{\infty} c_n x^n$ (where the $c_n$ are the Taylor series coefficients, $c_n = f^{(n)}(0)/n!$), you need to know that $R_n(x)$ (in the formula $f(x)=P_n(x)+R_n(x)$) tends to 0 as $n\to\infty.$ It is not sufficient to know that the series $\sum _{n=0} ^{\infty} = c_n x^n$ converges, because unfortunately it can sometimes happen that the series converges to a sum different from $f(x).$

The classic example of this is the function $f(x) = \begin{cases}e^{-1/x}&\text{if }x>0, \\ 0 &\text{if }x\leqslant 0.\end{cases}$ As shown here, that function has the property that its Taylor series about the point 0 converges to the identically zero function. But the function itself is not identically zero. What happens in that case is that all the coefficients in the Taylor series are 0. So in the formula $f(x)=P_n(x)+R_n(x)$, $P_n(x)$ is always the zero polynomial and $R_n(x)$ is always equal to $f(x).$ Thus $R_n(x)$ does not tend to 0 as $n\to\infty$, and hence $P_n(x)$ does not converge to $f(x).$

Can I also use the fact that the sum of continuous functions over a domain (in this case, $|x|<R$) is continuous, and that the sum of differentiable functions over a domain is differentiable?
Yes, those things are both true, provided that you mean a finite sum. If you allow infinite sums then those statements can both go wrong.
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
In order to be able to write $f(x)=\sum _{n=0} ^{\infty} c_n x^n$ (where the $c_n$ are the Taylor series coefficients, $c_n = f^{(n)}(0)/n!$), you need to know that $R_n(x)$ (in the formula $f(x)=P_n(x)+R_n(x)$) tends to 0 as $n\to\infty.$ It is not sufficient to know that the series $\sum _{n=0} ^{\infty} = c_n x^n$ converges, because unfortunately it can sometimes happen that the series converges to a sum different from $f(x).$

The classic example of this is the function $f(x) = \begin{cases}e^{-1/x}&\text{if }x>0, \\ 0 &\text{if }x\leqslant 0.\end{cases}$ As shown here, that function has the property that its Taylor series about the point 0 converges to the identically zero function. But the function itself is not identically zero...
I'm afraid I cannot agree with this statement!... As explained in...

Analytic Function -- from Wolfram MathWorld

... a function $f(z)$ is analytic in $z=z_{0}$ if and only if it is differentiable in $z=z_{0}$ and, because in this case $f(z)$ has derivatives of all order in $z=z_{0}$, its admits Taylor expansion in $z=z_{0}$ and the series converges to $f(z)$ ewerywhere inside a circle with center in $z_{0}$ and radius R. The 'classical example'...

$f(z) = \begin{cases}e^{-1/ z^{2}}&\text{if } z \ne 0, \\ 0 &\text{if } z=0 \end{cases}$ (1)

... doesn't have complex derivative in $z=0$ so that its Taylor expansion doesn't exist [in fact it exists the Laurent expansion around $z=0$...]. Regarding the 'mathematical material' contained in Wikipedia, in my opinion it is not all reliable and I strongly recommend 'Monster Wolfram'...

Kind regards

$\chi$ $\sigma$
 

OhMyMarkov

Member
Mar 5, 2012
83
Now I'm thinking, what if $c_n$ doesn't really represent coefficients "extracted" by Taylor's formula, what if $\sum c_n x^n$ is just an ordinary power series that converges for $x$ inside the radius. Then we do not have to worry about the remainder, given that the power series converges, right?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
I'm afraid I cannot agree with this statement!... As explained in...

Analytic Function -- from Wolfram MathWorld

... a function $f(z)$ is analytic in $z=z_{0}$ if and only if it is differentiable in $z=z_{0}$ and, because in this case $f(z)$ has derivatives of all order in $z=z_{0}$, its admits Taylor expansion in $z=z_{0}$ and the series converges to $f(z)$ ewerywhere inside a circle with center in $z_{0}$ and radius R. The 'classical example'...

$f(z) = \begin{cases}e^{-1/ z^{2}}&\text{if } z \ne 0, \\ 0 &\text{if } z=0 \end{cases}$ (1)

... doesn't have complex derivative in $z=0$ so that its Taylor expansion doesn't exist [in fact it exists the Laurent expansion around $z=0$...]. Regarding the 'mathematical material' contained in Wikipedia, in my opinion it is not all reliable and I strongly recommend 'Monster Wolfram'...

Kind regards

$\chi$ $\sigma$
There is of course a major difference between real differentiability and complex differentiability. The OP does not say whether the function in this thread is defined on the real or the complex numbers. If it defined on the complex numbers then it is true that a function that is differentiable at some point will always have a Taylor series in some neighbourhood of that point, and the Taylor series will converge to the function. I was answering the question on the assumption that it referred to a function of a real variable, where the situation is very different. As I explained above, a function on the real line can be (infinitely) differentiable there, with a convergent Taylor series whose sum is not equal to the function.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Now I'm thinking, what if $c_n$ doesn't really represent coefficients "extracted" by Taylor's formula, what if $\sum c_n x^n$ is just an ordinary power series that converges for $x$ inside the radius. Then we do not have to worry about the remainder, given that the power series converges, right?
Right!... it is important to write again what has been written in Your original post...

$\displaystyle f(x)= P_{n} (x) + R_{n} (x)$ (1)

... where $\displaystyle P_{n} (x) = \sum_{k=0}^{n} \frac{f^{(k)} (0)}{k!}\ x^{k}$ and $R_{n} (x)$ is a function that can assume some different look [Peano, Lagrange, Cauchy, etc...] but in any case tends to 0 if n tends to infinity. That shows clearly that to propose as 'counterexample' a Taylor expansion of f(x) that 'doesn't converge to f(x) [i.e. where $R_{n} (x)$ doesn't tend to 0 if n tends to infinity...] is at least innappropiate...

Kind regards

$\chi$ $\sigma$
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
A remarkable contribution to that argument comes from an old our 'acquaitance'... Pauls Online Notes : Calculus II - Taylor Series

We report...

To determine a condition that must be true in order for a Taylor series to exist for a function let’s first define the nth degree Taylor polynomial of $f(x)$ as...


$\displaystyle T_{n}= \sum_{i=0}^{n} \frac{f^{(i)} (a)}{i!}\ (x-a)^{i}$ (1)

Next the remainder is defined as...

$\displaystyle R_{n} (x) = f(x) - T_{n} (x)$

... so that...

$\displaystyle f(x) = T_{n} (x) + R_{n} (x)$ (2)

We have now the following ...

Theorem

Suppose the (2) is true, then if...

$\displaystyle \lim_{n \rightarrow \infty} R_{n} (x)=0$

... for $|x-a|<r$, then...

$\displaystyle f(x)= \sum_{i=0}^{\infty} \frac{f^{(i)} (a)}{i!}\ (x-a)^{i}$

... for $|x-a|<r$.

Kind regards

$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Now I'm thinking, what if $c_n$ doesn't really represent coefficients "extracted" by Taylor's formula, what if $\sum c_n x^n$ is just an ordinary power series that converges for $x$ inside the radius. Then we do not have to worry about the remainder, given that the power series converges, right?
That is correct. If a function is defined by a power series, $f(x) = \sum c_nx^n$ then the pathology I mentioned in previous comments cannot occur. So long as you stay inside the radius of convergence, the series will converge to the function. What is more the function will be differentiable, and you can find its derivative by differentiating the power series term by term, $f'(x) = \sum nc_nx^{n-1}$, and the power series for $f'(x)$ will have the same radius of convergence as the power series for $f(x).$