Re: taylor expansions with real vectors

[bilelsan]

Dear list,

I have a big deal concerning the development of a Taylor expansion.

require(Matrix)
e1 <- as.vector(1:5)
e2 <- as.vector(6:10)

in order to obtain all the combinations between these two vectors following
a Taylor expansion (or more simply through a Maclaurin series) for real
numbers.
We have f(x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + … + f^(k)(0)(x-0)^k/k!
with
f(x) = e1 + e2 for Taylor expansion (r = 1)
        + 1/2!*e1^2 + 1/2!*e2^2 + 1/2!*e1*e2 for Taylor expansion (r = 2)
excluding e2*e1
        + 1/3!*e1^3 + 1/3!*e1^2*e2 + 1/3!*e2^2*e1 + 1/3!*e2^3 for Taylor
expansion (r = 3) excluding e2*e1^2 and e1*e2^2
       ...
I already write the number of possible combinations as :
x <- as.vector(0)
for (r in 1:r){x[r] <- 2*(sum(choose(2*q+r-1,r))-sum(choose(q+r-1,r)))}# q:
number of lag of e1 and e2; r: order of taylor expansion
nstar   <- sum(x) # N* number of total combinations
 
How to write f(x) in a general framework?
Quid of this framework when e1 and e2 are completed with their lags if q >
1?
Your help or advice would be greatly appreciated

Bilel

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[Gabor Grothendieck]

On Wed, Jul 18, 2012 at 9:47 AM, bilelsan <[hidden email]> wrote:

> Dear list,
>
> I have a big deal concerning the development of a Taylor expansion.
>
> require(Matrix)
> e1 <- as.vector(1:5)
> e2 <- as.vector(6:10)
>
> in order to obtain all the combinations between these two vectors following
> a Taylor expansion (or more simply through a Maclaurin series) for real
> numbers.
> We have f(x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + … + f^(k)(0)(x-0)^k/k!
> with
> f(x) = e1 + e2 for Taylor expansion (r = 1)
>         + 1/2!*e1^2 + 1/2!*e2^2 + 1/2!*e1*e2 for Taylor expansion (r = 2)
> excluding e2*e1
>         + 1/3!*e1^3 + 1/3!*e1^2*e2 + 1/3!*e2^2*e1 + 1/3!*e2^3 for Taylor
> expansion (r = 3) excluding e2*e1^2 and e1*e2^2
>        ...
> I already write the number of possible combinations as :
> x <- as.vector(0)
> for (r in 1:r){x[r] <- 2*(sum(choose(2*q+r-1,r))-sum(choose(q+r-1,r)))}# q:
> number of lag of e1 and e2; r: order of taylor expansion
> nstar   <- sum(x) # N* number of total combinations
>
> How to write f(x) in a general framework?
> Quid of this framework when e1 and e2 are completed with their lags if q >
> 1?
> Your help or advice would be greatly appreciated
>

See the section on Taylor expansions in the Ryacas package vignette.
Depending on what you want to do that may or may not be relevant.

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GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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[ruseraix]

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[Petr Savicky]

On Wed, Jul 18, 2012 at 06:02:27PM -0700, bilelsan wrote:
> Leave the Taylor expansion aside, how is it possible to compute with [R]:
> f(e) = e1 + e2 #for r = 1
> + 1/2!*e1^2 + 1/2!*e2^2 + 1/2!*e1*e2 #for r = 2, excluding e2*e1
> + 1/3!*e1^3 + 1/3!*e1^2*e2 + 1/3!*e2^2*e1 + 1/3!*e2^3 #for r = 3, excluding
> e2*e1^2 and e1*e2^2
> + ... #for r = k
> In other words, I am trying to figure out how to compute all the possible
> combinations as exposed above.

Hi.

For a general r, do you mean the following sum of products?

  1/r! (e1^r + e1^(r-1) e2 + ... e1 e2^(r-1) + e2^r)

If this is correct, then try

  f <- 0
  for (r in 1:k) {
     f <- f + 1/factorial(r) * sum(e1^(r:0)*e2^(0:r))
  }

Hope this helps.

Petr Savicky.

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[ruseraix]

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[Petr Savicky]

On Sun, Jul 22, 2012 at 05:34:08PM -0700, bilelsan wrote:
> Hi,
> Thanks a lot for answer. It is what I mean.
> But the code does not seem to work (

Hi.

I am sorry for a late reply. I was on vacations one week.

Can you specify the problem?

I get

  e1 <- 2
  e2 <- 3
  k <- 3
  f <- 0
  for (r in 1:k) {
     f <- f + 1/factorial(r) * sum(e1^(r:0)*e2^(0:r))
  }
  f

  [1] 25.33333

If this the correct answer?

The code may be put to a function.

  getF <- function(e1, e2, k)
  {
      f <- 0
      for (r in 1:k) {
          f <- f + 1/factorial(r) * sum(e1^(r:0)*e2^(0:r))
      }
      f
  }

  getF(2, 3, 3)

  [1] 25.33333

Hope this helps.

Petr Savicky.

>
> Le Jul 19, 2012 ?? 8:52 AM, Petr Savicky [via R] a ??crit :
>
> > On Wed, Jul 18, 2012 at 06:02:27PM -0700, bilelsan wrote:
> > > Leave the Taylor expansion aside, how is it possible to compute with [R]:
> > > f(e) = e1 + e2 #for r = 1
> > > + 1/2!*e1^2 + 1/2!*e2^2 + 1/2!*e1*e2 #for r = 2, excluding e2*e1
> > > + 1/3!*e1^3 + 1/3!*e1^2*e2 + 1/3!*e2^2*e1 + 1/3!*e2^3 #for r = 3, excluding
> > > e2*e1^2 and e1*e2^2
> > > + ... #for r = k
> > > In other words, I am trying to figure out how to compute all the possible
> > > combinations as exposed above.
> >
> > Hi.
> >
> > For a general r, do you mean the following sum of products?
> >
> >   1/r! (e1^r + e1^(r-1) e2 + ... e1 e2^(r-1) + e2^r)
> >
> > If this is correct, then try
> >
> >   f <- 0
> >   for (r in 1:k) {
> >      f <- f + 1/factorial(r) * sum(e1^(r:0)*e2^(0:r))
> >   }
> >
> > Hope this helps.
> >
> > Petr Savicky.
> >
> > ______________________________________________
> > [hidden email] mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
> >
> > If you reply to this email, your message will be added to the discussion below:
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> > NAML
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>
>
>
>
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