Tilted Forum Project Discussion Community - View Single Post - Rapidly finding terms of Taylor polynomial

phukraut · 04-11-2005, 09:53 AM

Looking at things a different way, we see a pattern. Let aij 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>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>0).

Let Sm,p = a0ps0 + a1ps1 + ... + ampsm, 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.

04-11-2005, 09:53 AM	#9 (permalink)
phukraut Addict	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>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>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..