Rapidly finding terms of Taylor polynomial
 

w = INTEGRAL e^sin(t) dt

w has no indefinite form. But a Taylor polynomial can be generated for

w(x) = INTEGRAL₀^x e^sin(t) dt

Where it is centered around zero. However, to do this rapidly there needs to be a quick way to determine the nth derivative of w(x).
I have noted that the nth derivative always has the term e^sin(t), so I have started with letting the nth derivative be equal to:

w⁽ⁿ⁾(x) = f_n(x)*e^sin(t)

Where:

f₁(x) = 1 ;
f₂(x) = cos(x) ;
f₃(x) = cos²(x) - sin(x) ; so on and so forth if you wish to find each derivative.

If the derivative of w(x) continues then the pattern that f_n(x) follows is:

f_n(x) = f_n-1(x)*cos(x) + ^d/_dx f_n-1(x)

Because the Taylor function we desire is centered at 0 let:

g(n) = f_n(0)

The Taylor polynomial then becomes

SUM_n=1^T g(n)*xⁿ/n!

My question is: what is g(n) so that a single summation formula can be made to describe INTEGRAL₀^x e^sin(t) dt?

I'm don't know the answer to your question, but you can use the SUP and SUB html tags for superscript and subscript. i.e.
X[sup]y[/sup] gives: X^y

when you replace the [] with <>
and

X[sub]n[/sub] gives: X_n

you can even nest the tags if necessary:
X_n^y_i²

Sorry, the original version of my post didn't make any sense and I don't have time right now to make a new one that's helpful, but hopefully I'll figure something out later. ;)

What I am trying to do is create a function that gives the numerical value of the nth derivative of e^sin(t) at zero to create a single sum formula for the integral of that equation.

EDIT: Sorry, didn't see that. I really just woke up and am not banging on all cylinders as it were. Yes, it's 5 in the afternoon but I have been busy and have not had my morning dose of caffeine.

No, you misunderstand, I meant the original version of MY post didn't make sense and I edited it to say what it does now.

I don't have the answer, but to provide some data, here is the expansion of the derivatives of exp(sin(0)) for n=0,1,2,...,15:

1, 1, 1, 0, -3, -8, -3, 56, 217, 64, -2951, -12672, 5973, 309376, 1237173, -2917888.

The OEIS gives <a href="http://www.research.att.com/projects/OEIS?Anum=A002017">A002017</a>, which doesn't provide a general formula unfortunately.

If, in f_n(t), we let s=sin(t) and c=cos(t), then the expansion looks like this:

f₀(t) = 1s⁰c⁰.

f₁(t) = 1s⁰c¹.

f₂(t) = -1s¹c⁰ + 1s⁰c².

f₃(t) = -1s⁰c¹ - 3s¹c¹ + 1s⁰c³.

And so on. The coefficients here are trinomials. If I get any further I'll post here.

Phukraut, your system that you posted looks promising. But f₁(x) = 1 by the way I am looking at w⁽ⁿ⁾(x).

w⁽ⁿ⁾(x) = ^{d^n-1}/_dx^n-1 e^sin(x)

So e^sin(x) is th first deriviative.

Right. I'm not exactly sure why you've made the offset 1 instead of 0, but in the end the result is easily modified to account for this difference.

Looking at things a different way, we see a pattern. Let a_ij be the coefficient in my above list for the term with sine to the power of i and cosine to the power of j---so for example, a₃₂s³c². Further, to avoid a later problem with notation, let n&gt;0. Now look at the values of f₆ and f₇ when we factor out cosine.

f₆(t) = c⁰(a₁₀s¹ + a₂₀s² + a₃₀s³) + c²(a₀₂s⁰ + a₁₂s¹ + a₂₂s²) + c⁴(a₀₄s⁰ + a₁₄s¹) + c⁶(a₀₆s⁰).

f₇(t) = c¹(a₀₁s⁰ + a₁₁s¹ + a₂₁s² + a₃₁s³) + c³(a₀₃s⁰ + a₁₃s¹ + a₂₃s²) + c⁵(a₀₅s⁰ + a₁₅s¹) + c⁷(a₀₇s⁰).

I believe I have the pattern (if we let n&gt;0).

Let S_m,p = a_0ps⁰ + a_1ps¹ + ... + a_mps^m, where m,p=0,1,2,.... Then we split n into even and odd terms and look at each pattern separately.

Consider k=1,2,3,..., and n=2k. Then we have

http://img17.paintedover.com/uploads/17/even3.gif

Then for odd n, (n=2k+1, where k=0,1,2,...), we see that

http://img17.paintedover.com/uploads/17/odd2.gif

I hope I didn't make any mistakes there. Anyway, I'll try and combine the two later.

Taking t=0, we have

http://img17.paintedover.com/uploads/17/evensimp2.gif

And for odd,

http://img17.paintedover.com/uploads/17/oddsimp.gif

Now all that's left is to actually find the coefficients.

Since sin(0)=0, we can set a_ij=0 for all i&gt;0, or simply disregard these coefficients. So here are some of the coefficients apparently solved as a_0j (noting that binomial(i,j) is the binomial coefficient for integers i and j where, if i&lt;j, binomial(i,j)=0):

a_0n = 1 for all n;
a_0(n-2) = (-1)binomial(n,3) for all n;
a_0(n-4) = (1/3)(5n - 22)binomial(n,5) for all n&gt;4.

I am blinking but it is starting to sink in. I have never dealt with this form of notation before. And the reason I am "offsetting" my derivatives is because the first derivative of w(x) = e^sin(x).

This is just one approach. Another is to develop your differential above recursively. You may have more luck with that, since I'm currently stuch on the constants a_0(n-6), which begin (at n=7),

-1, -64, -820, -5440, -24970, -90112, -458172, -1036035, -1768195, -3935360, ....

Cheers.