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Old 04-09-2005, 04:36 PM   #1 (permalink)
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Rapidly finding terms of Taylor polynomial

w = INTEGRAL e<sup>sin(t)</sup> dt
w has no indefinite form. But a Taylor polynomial can be generated for
w(x) = INTEGRAL<sub>0</sub><sup>x</sup> e<sup>sin(t)</sup> 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<sup>sin(t)</sup>, so I have started with letting the nth derivative be equal to:
w<sup>(n)</sup>(x) = f<sub>n</sub>(x)*e<sup>sin(t)</sup>
Where:
f<sub>1</sub>(x) = 1 ;
f<sub>2</sub>(x) = cos(x) ;
f<sub>3</sub>(x) = cos<sup>2</sup>(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<sub>n</sub>(x) follows is:
f<sub>n</sub>(x) = f<sub>n-1</sub>(x)*cos(x) + <sup>d</sup>/<sub>dx</sub> f<sub>n-1</sub>(x)
Because the Taylor function we desire is centered at 0 let:
g(n) = f<sub>n</sub>(0)
The Taylor polynomial then becomes
SUM<sub>n=1</sub><sup>T</sup> g(n)*x<sup>n</sup>/n!
My question is: what is g(n) so that a single summation formula can be made to describe INTEGRAL<sub>0</sub><sup>x</sup> e<sup>sin(t)</sup> dt?
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Last edited by Hain; 04-11-2005 at 07:55 AM.. Reason: Thank you CSFlim
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Old 04-10-2005, 07:58 AM   #2 (permalink)
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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<sup>y</sup>

when you replace the [] with <>
and

X[sub]n[/sub] gives: X<sub>n</sub>

you can even nest the tags if necessary:
X<sub>n</sub><sup>y<sub>i</sub><sup>2</sup></sup>
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Old 04-10-2005, 10:43 AM   #3 (permalink)
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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.
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Last edited by n0nsensical; 04-10-2005 at 01:52 PM..
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Old 04-10-2005, 01:45 PM   #4 (permalink)
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What I am trying to do is create a function that gives the numerical value of the nth derivative of e<sup>sin(t)</sup> 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.
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Last edited by Hain; 04-10-2005 at 01:55 PM..
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Old 04-10-2005, 01:52 PM   #5 (permalink)
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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.
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"Prohibition will work great injury to the cause of temperance. It is a species of intemperance within itself, for it goes beyond the bounds of reason in that it attempts to control a man's appetite by legislation, and makes a crime out of things that are not crimes. A Prohibition law strikes a blow at the very principles upon which our government was founded." --Abraham Lincoln
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Old 04-10-2005, 02:31 PM   #6 (permalink)
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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<sub>n</sub>(t), we let s=sin(t) and c=cos(t), then the expansion looks like this:

f<sub>0</sub>(t) = 1s<sup>0</sup>c<sup>0</sup>.

f<sub>1</sub>(t) = 1s<sup>0</sup>c<sup>1</sup>.

f<sub>2</sub>(t) = -1s<sup>1</sup>c<sup>0</sup> + 1s<sup>0</sup>c<sup>2</sup>.

f<sub>3</sub>(t) = -1s<sup>0</sup>c<sup>1</sup> - 3s<sup>1</sup>c<sup>1</sup> + 1s<sup>0</sup>c<sup>3</sup>.

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

Last edited by phukraut; 04-10-2005 at 02:44 PM..
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Old 04-11-2005, 08:11 AM   #7 (permalink)
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Phukraut, your system that you posted looks promising. But f<sub>1</sub>(x) = 1 by the way I am looking at w<sup>(n)</sup>(x).
w<sup>(n)</sup>(x) = <sup>d<sup>n-1</sup></sup>/<sub>dx<sup>n-1</sup></sub> e<sup>sin(x)</sup>
So e^sin(x) is th first deriviative.
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Last edited by Hain; 04-11-2005 at 06:34 PM..
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Old 04-11-2005, 08:49 AM   #8 (permalink)
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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.
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Old 04-11-2005, 09:53 AM   #9 (permalink)
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Looking at things a different way, we see a pattern. Let a<sub>ij</sub> 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, a<sub>32</sub>s<sup>3</sup>c<sup>2</sup>. Further, to avoid a later problem with notation, let n&gt;0. Now look at the values of f<sub>6</sub> and f<sub>7</sub> when we factor out cosine.

f<sub>6</sub>(t) = c<sup>0</sup>(a<sub>10</sub>s<sup>1</sup> + a<sub>20</sub>s<sup>2</sup> + a<sub>30</sub>s<sup>3</sup>) + c<sup>2</sup>(a<sub>02</sub>s<sup>0</sup> + a<sub>12</sub>s<sup>1</sup> + a<sub>22</sub>s<sup>2</sup>) + c<sup>4</sup>(a<sub>04</sub>s<sup>0</sup> + a<sub>14</sub>s<sup>1</sup>) + c<sup>6</sup>(a<sub>06</sub>s<sup>0</sup>).

f<sub>7</sub>(t) = c<sup>1</sup>(a<sub>01</sub>s<sup>0</sup> + a<sub>11</sub>s<sup>1</sup> + a<sub>21</sub>s<sup>2</sup> + a<sub>31</sub>s<sup>3</sup>) + c<sup>3</sup>(a<sub>03</sub>s<sup>0</sup> + a<sub>13</sub>s<sup>1</sup> + a<sub>23</sub>s<sup>2</sup>) + c<sup>5</sup>(a<sub>05</sub>s<sup>0</sup> + a<sub>15</sub>s<sup>1</sup>) + c<sup>7</sup>(a<sub>07</sub>s<sup>0</sup>).

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

Let S<sub>m,p</sub> = a<sub>0p</sub>s<sup>0</sup> + a<sub>1p</sub>s<sup>1</sup> + ... + a<sub>mp</sub>s<sup>m</sup>, 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



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



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

Last edited by phukraut; 04-11-2005 at 11:45 AM..
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Old 04-11-2005, 11:50 AM   #10 (permalink)
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Taking t=0, we have



And for odd,



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

Last edited by phukraut; 04-11-2005 at 12:28 PM..
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Old 04-11-2005, 01:42 PM   #11 (permalink)
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Since sin(0)=0, we can set a<sub>ij</sub>=0 for all i&gt;0, or simply disregard these coefficients. So here are some of the coefficients apparently solved as a<sub>0j</sub> (noting that binomial(i,j) is the binomial coefficient for integers i and j where, if i&lt;j, binomial(i,j)=0):

a<sub>0n</sub> = 1 for all n;
a<sub>0(n-2)</sub> = (-1)binomial(n,3) for all n;
a<sub>0(n-4)</sub> = (1/3)(5n - 22)binomial(n,5) for all n&gt;4.

Last edited by phukraut; 04-11-2005 at 03:19 PM.. Reason: Correction on the a<sub>ij</sub>=0. This is in fact not true. What I wanted to say was that we can disregard these coefficients since the term is identically zero.
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Old 04-11-2005, 02:38 PM   #12 (permalink)
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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).
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Old 04-11-2005, 03:10 PM   #13 (permalink)
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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<sub>0(n-6)</sub>, which begin (at n=7),

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

Cheers.

Last edited by phukraut; 04-11-2005 at 03:20 PM..
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