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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2012-11-30 14:38:17

Hi! smile

The third degree polynomial divided by the fifth degree polynomial can be written as an infinite
series as:

Then just integrate it term by term to get the indefinite integral in an infinite series form.

Starting with the 1/x^2 term the coefficients in the series repeat in blocks of 10 in the pattern
1,-1,1,0,0,-1,1,-1,0,0.  That is what the 1/(x^10n) handles.  So the next 6 terms in the series
would have exponents on the bottom of 12,13,14,17,18,19 and the next 6 would have
22,23,24,27,28,29 for the exponents on the bottom.

The signs in each block of six terms would continue to be 1, -1, 1, -1, 1, -1.

Has anyone got a closed form for the integral?  I'd love to see it.  If x^5+1 is divided by x-1 I get
x(x+1)(x-1)(x^2+1)  +  (x+1)/(x-1)  but I haven't been able to use this to get a closed form.

Good luck with it! dizzy

2012-11-30 08:20:54


Yes, it looks like a handmade one. That can be dangerous.

2012-11-30 05:29:25

The integral can be solved through partial fractions, but it is a nasty factorisation!

bob bundy
2012-11-30 03:26:55

Sorry, you are right.  Selective blindness again.

I was having trouble with my minus signs. 


2012-11-30 02:19:20

It is a factor of the denominator only.

2012-11-30 00:46:46

bob bundy wrote:

(x + 1) is a factor of top and bottom.


It is not.

2012-11-29 20:15:16

Hi jacks;

The answer for that is huge, are you supposed to do that by hand?

bob bundy
2012-11-29 20:00:08

(x + 1) is a factor of top and bottom.


2012-11-29 19:56:22

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