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Question: How to convert to sum?

Question:How to convert to sum?

Ronan 1207 Maple 2018
sum eval series limit
August 09 2018
1

I am trying to evaluate some functions when u=v so function goes to 0/0

Example given below of the expansion series I need to produce first. But I can't produce it.
 

restart

"How do I get this expansion and sum"?""

````

``

eq := (u^m-v^m)/(u-v)

(u^m-v^m)/(u-v)

 (1)

eval(eq, u = v)

Error, numeric exception: division by zero

  

``

eq1 := convert(eq, Sum, method = rational, include = powers)

(Sum((1-u)^_k1*binomial(-m+_k1-1, _k1), _k1 = 0 .. m)-(Sum((1-v)^_k1*binomial(-m+_k1-1, _k1), _k1 = 0 .. m)))*(Sum((-1)^_k1*(-1+u-v)^_k1, _k1 = 0 .. infinity))

 (2)

"(=)"

-(Sum((-1)^_k1*(-1+u-v)^_k1, _k1 = 0 .. infinity))*m*(Sum((1-u)^_k1*binomial(-m+_k1, _k1)/(-m+_k1), _k1 = 0 .. m))+(Sum((-1)^_k1*(-1+u-v)^_k1, _k1 = 0 .. infinity))*m*(Sum((1-v)^_k1*binomial(-m+_k1, _k1)/(-m+_k1), _k1 = 0 .. m))

 (3)

"(=)"

-(Sum((-1)^_k1*(-1+u-v)^_k1, _k1 = 0 .. infinity))*m*(Sum((1-u)^_k1*binomial(-m+_k1, _k1)/(-m+_k1), _k1 = 0 .. m))+(Sum((-1)^_k1*(-1+u-v)^_k1, _k1 = 0 .. infinity))*m*(Sum((1-v)^_k1*binomial(-m+_k1, _k1)/(-m+_k1), _k1 = 0 .. m))

 (4)

eval(eq1, u = v)

0

 (5)

for m to 5 do m, " ", simplify(eq); eval(simplify(eq), u = v) end do

5*v^4

 (6)

unassign('m')

``eq2 := sum(u^(m-1-i)*v^i, i = 0 .. m-1)

-u^(m-1)*(v/u)^m*u/(u-v)+u^(m-1)*u/(u-v)

 (7)

"(=)"

-u^m*((v/u)^m-1)/(u-v)

 (8)

eval(eq2, u = v)

Error, numeric exception: division by zero

  

``

``

u := v

v

 (9)

eq3 := sum(u^(m-1-i)*v^i, i = 0 .. m-1)

v^m*m/v

 (10)

"(=)"

v^(m-1)*m

 (11)

``


 

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