How Can We Ensure Convergence in Function Approximations Beyond Taylor Series?

[mertcan]

Hi, as you know infinite sum of taylor series may not converge to its original function which means when we increase the degree of series then we may diverge more. Also you know taylor series is widely used for an approximation to vicinity of relevant point for any function. Let's think about a function which is infinitely differentiable so infinitely taylor series exist but it is so complicated function that we can not apply ratio test to find radius of convergence or integral test analytically...In short how we can approximate to any function with a guaranteed convergence except taylor series expansion?

 

[mathman]

There are no guarantees.

 

[BvU]

as you know infinite sum of taylor series may not converge to its original function

Enlighten us ...

Let's think about a function which is infinitely differentiable so infinitely taylor series exist but it is so complicated function that we can not apply ratio test to find radius of convergence or integral test analytically

Can you give an example ? Is there something specific you have in mind ?

 

[scottdave]

I'm curious. Is this a homework question, or just something to ponder?

 

[mertcan]

Hi, I would like to express: Taylor series expansion is widely used in Finite Element methods for instance to set the relation between nodes. I agree to use Taylor expansion in finite difference schemes to approximate to a node from another node but I have not seen any case that Taylor series used in schemes are controlled in terms of convergence. Taylor expansion may converge or not to its original function which means you may diverge more when you increase the degree of Taylor series expansion but in FEM it is always said that Taylor expansion gets more accurate and thus scheme gets more precise when expansion degree increases although there is not a convergence check for taylor expansion. Could you enlighten me about that? Why convergence of taylor expansion between nodes in finite difference schemes are not checked?

 

[anorlunda]

[Moderator note: The OP reprhased his question, and created a second thread. I merged that back into this thread.]

@mertcan , Please think about and respond to the replies you have rather than ignore them and repost the same question.

 

[mertcan]

@BvU @mathman @scottdave what are your thoughts about my last question which also involve Finite element method?

 

[mathman]

@BvU @mathman @scottdave what are your thoughts about my last question which also involve Finite element method?

Sorry: I have had no experience with FEM so I can't give any useful comment here.

 

[anorlunda]

[Moderator note: this thread was moved here from the Calculus forum.]

 

[mertcan]

Let me ask my question in post 5 in that way: If we expand Taylor series of infinitely differentiable function then we may diverge because of the fact that derivative at infinity may be infinite or unbounded. Convergence tests I have seen so far are always capable of reflecting the formula of coefficients of power series at infinity in short we know what the coefficients will be at infinity. Taylor series could be considered also as a power series whose coefficients are derivatives of function but we may not know the formula of derivatives (coefficients of power series) at infinity. So how can we find out the radius of convergence of relevant function whose derivatives are not known at infinity in Taylor series expansion?

 

[pbuk]

Are you asking whether there exist differential equations for which higher order numerical methods do not perform well?

Yes this is certainly true.

In short how we can approximate to any function with a guaranteed convergence

We can't. There are some functions which are difficult to integrate numerically,

in FEM it is always said that Taylor expansion gets more accurate and thus scheme gets more precise when expansion degree increases

Who always says this? I do not agree with them. Sometimes a higher-order method is better, sometimes you are better off with smaller steps, sometimes round-off error is the limiting factor and you need to use other methods...

 

[mertcan]

Are you asking whether there exist differential equations for which higher order numerical methods do not perform well?

Yes this is certainly true.We can't. There are some functions which are difficult to integrate numerically,Who always says this? I do not agree with them. Sometimes a higher-order method is better, sometimes you are better off with smaller steps, sometimes round-off error is the limiting factor and you need to use other methods...

Thans for return @pbuk. Could I get your valuable response to my following question also I mentioned in post 10 :Taylor series could be considered also as a power series whose coefficients are derivatives of function but we may not know the formula of derivatives (coefficients of power series) at infinity. So how can we find out the radius of convergence of relevant function whose derivatives are not known at infinity in Taylor series expansion?

 

[pbuk]

So how can we find out the radius of convergence of relevant function whose derivatives are not known at infinity in Taylor series expansion?

We can't, so we use other methods to estimate the accuracy of our numerical method, or detect if it is unstable.

 

[mertcan]

We can't, so we use other methods to estimate the accuracy of our numerical method, or detect if it is unstable.

Thanks for response...I have 2 things to say @pbuk: 1) what methods you mentioned are employed to estimate the accuracy of our numerical method?
2) As far as I know, Von neumann stability analysis or other similar stability analysis consider the explosiveness of finite difference scheme just with respect to time( to prevent divergence situation just over time not grid size...). If we think of that time is frozen and only coordinate or grid size terms are left then scheme may diverge because of divergence of Taylor expansion. What do you say?

 

[pbuk]

1) what methods you mentioned are employed to estimate the accuracy of our numerical method?

I think this is my last post on this topic because you should be able to find out answers in whatever sources you are using to learn from, but here we go with some google searches as a substitute...
https://www.google.com/search?q=boundary+value+problem+error+estimation

2) As far as I know, Von neumann stability analysis or other similar stability analysis consider the explosiveness of finite difference scheme just with respect to time( to prevent divergence situation just over time not grid size...).

https://www.google.com/search?q=von+neumann+stability+analysis+finite+difference+methodsOr this one looks like 185 pages of good stuff: https://people.maths.ox.ac.uk/trefethen/4all.pdf
Or Chapters 8-10 of Numerical Computation of Internal External Flows Vol 1 (Hirsch) - this seems to be available on the web but I am not sure about copyright so I won't provide a link.

 

[mertcan]

I think this is my last post on this topic because you should be able to find out answers in whatever sources you are using to learn from, but here we go with some google searches as a substitute...
https://www.google.com/search?q=boundary+value+problem+error+estimationhttps://www.google.com/search?q=von+neumann+stability+analysis+finite+difference+methodsOr this one looks like 185 pages of good stuff: https://people.maths.ox.ac.uk/trefethen/4all.pdf
Or Chapters 8-10 of Numerical Computation of Internal External Flows Vol 1 (Hirsch) - this seems to be available on the web but I am not sure about copyright so I won't provide a link.

Thanks @pbuk,I would like to ask a little question about stability analysis: In order to ensure von neumann stability analysis, we may have $$a*\frac {\Delta_t} {\Delta_x} < 1$$ for a finite difference scheme whereas a is constant. I deem that even if we provide stability analysis our solution may not be consistent in terms of taylor expansion. For instance say that $$\Delta_t <<<<<<<< 1$$ and $$\Delta_x = 10000000$$ Here stability is ensured but due to the fact that grid size ,which means Delta_x, is so large than taylor expansion tends to diverge. In short stability results are not useful to obtain consistent or convergent taylor expanison therefore what else can we do to have right grid size which lead to convergent taylor expanison?

 

[mertcan]

Thanks @pbuk,I would like to ask a little question about stability analysis: In order to ensure von neumann stability analysis, we may have $$a*\frac {\Delta_t} {\Delta_x} < 1$$ for a finite difference scheme whereas a is constant. I deem that even if we provide stability analysis our solution may not be consistent in terms of taylor expansion. For instance say that $$\Delta_t <<<<<<<< 1$$ and $$\Delta_x = 10000000$$ Here stability is ensured but due to the fact that grid size ,which means Delta_x, is so large than taylor expansion tends to diverge. In short stability results are not useful to obtain consistent or convergent taylor expanison therefore what else can we do to have right grid size which lead to convergent taylor expanison?

@pbuk what do you consider about my last post?

 

[pbuk]

@pbuk what do you consider about my last post?

What do you consider about your last post? If you can learn about $\Delta t$ by investigating the partial deriviative with respect to $t$, how do you think you can learn about $\Delta x$?

 

[mertcan]

What do you consider about your last post? If you can learn about $\Delta t$ by investigating the partial deriviative with respect to $t$, how do you think you can learn about $\Delta x$?

@pbuk I just would like to emphasise that stability analysis results mostly gives an interval including $\Delta t$ and $\Delta x$ . And I deem that in FEM we try to have stable system according to values which obey the interval, so as long as we have consistency with stability I consider we can choose whatever the $\Delta x$ or $\Delta t$ is. But Even if we have stability we may diverge in terms of taylor series right? Could you help me about that confusion?

 

[mertcan]

I always see that taylor expansion usage is so pervasive in finite element/difference method. Actually it implies that if taylor expansion is employed then it must be accepted that all derivatives of function need to be bounded to be convergent to original function. What should be done in FEM or FDM if derivatives of function may be unbounded? or Is it possible to have a function in FEM or FDM with unbounded derivative?

 

[mertcan]

I always see that taylor expansion usage is so pervasive in finite element/difference method. Actually it implies that if taylor expansion is employed then it must be accepted that all derivatives of function need to be bounded to be convergent to original function. What should be done in FEM or FDM if derivatives of function may be unbounded? or Is it possible to have a function in FEM or FDM with unbounded derivative?

@pbuk also May I get your valuable comments?

 

[FactChecker]

The convergence of Taylor series can only be well understood when the complex plane is considered. When they converge, their behavior is very good and it possesses desirable properties. There are other series, like Fourier series, which converge for more functions on a segment of the real line. And there are integral representations that are useful for even more functions on the real line (see Fourier inversion theorem ).

 

[pbuk]

@pbuk also May I get your valuable comments?

What should be done in FEM or FDM if derivatives of function may be unbounded?

Use a smaller grid.

This thread is going in circles.

 

[fresh_42]

This thread is going in circles.

Thread closed. Thanks for your participation.