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Math Forum » Discussions » sci.math.* » sci.math.symbolic.independent

Topic: It's teatime at the End of the Universe ...
Replies: 2   Last Post: Oct 23, 2011 8:03 AM

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clicliclic@freenet.de

Posts: 961
Registered: 4/26/08
It's teatime at the End of the Universe ...
Posted: Oct 15, 2011 4:09 AM
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... today the Restaurant at the End of the Universe is offering free
cookies made by the chef ... fresh snaaacks! ... free cookies made by
the cheeef! ... you look like you want one? ... help yourself, Sir ...

(m + 1)*(a*d - b*c)*(a*f - b*e)
*INT((A + B*x)*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x)
- INT(((m + 1)*(A*(a*d*f - (c*f + d*e)*b) + B*b*c*e) - (A*b - B*a)
*((n + 1)*d*e + (p + 1)*c*f) - (m + n + p + 3)*(A*b - B*a)*d*f*x)
*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x)
- (A*b - B*a)
*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1) = 0

(m + n + p + 2)*d*f
*INT((A + B*x)*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x)
- INT(((m + n + p + 2)*A*a*d*f - B*(m*b*c*e
+ ((n + 1)*d*e + (p + 1)*c*f)*a) + ((m + n + p + 2)*A*b*d*f
+ B*(m*a*d*f - ((m + n + 1)*e*d + (m + p + 1)*c*f)*b))*x)
*(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p, x)
- B*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1) = 0

(m + 1)*(a*f - b*e)*b
*INT((A + B*x)*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x)
- INT(((m + 1)*(A*f - B*e)*b*c + (A*b - B*a)*(n*d*e + (p + 1)*c*f)
+ ((m + 1)*(A*f - B*e)*b + (n + p + 1)*(A*b - B*a)*f)*d*x)
*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x)
+ (A*b - B*a)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1) = 0

... they say you are aiming to be the first to draw a three-dimensional
domain-map - that's something! ... fresh snaaacks! ... free cookies
made by the cheeef! .... Want one too? ... note that the next four
require a*d - b*c = 0 ...

(m + n + 1)*(a*f - b*e)*INT((a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x)
- (m + n + p + 2)*f*INT((a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x)
+ (a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1) = 0

(p + 1)*(c*f - d*e)*INT((a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x)
+ (m + n + p + 2)*d*INT((a + b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1), x)
- (a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1) = 0

(m + n + 1)*b*INT((a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x)
+ p*f*INT((a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p - 1), x)
- (a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p = 0

b*INT((a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x)
- d*INT((a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x) = 0

... that last one looks especially yummy ... unfortunately we have only
one today ... fresh snaaacks! ... free cookies made by the cheeef! ...
hungry for a snack? ... the next two require a*d - b*c = c*f - d*e = 0
...

(m + n + p + 1)*f*INT((a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x)
- (a + b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1) = 0

f*INT((a + b*x)^m*(c + d*x)^n/(e + f*x)^(m + n + 1), x)
- (a + b*x)^m*(c + d*x)^n/(e + f*x)^(m + n)*LN(e + f*x) = 0

... nonrecursive creations are a real strength of our chef ... fresh
snaaacks! ... free cookies at the End of the Universe ...

Martin.




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