50G How to transform '1/(a+1)*X*X^a' into '1/(a+1)*X^(a+1)' }

[Gil]

For the following, please excuse my lack of rigorous (mathematical) reasonings, as I haven't got a scientific background.

Here I venture.

After second thought, it may now seem logical to me that the HP50G calculator does not simplify 'X * X^a' (with any command, but in particular with SIMPLIFY command or EXPAND command), but does simplify an expression of the type 'X * X^2', with a number (here 2, in that illustrative case) instead of the variable a (with again first mentioned command SIMPLIFY).

A)
Consider X>=0.
Then X=EXP(LN(X) <0>
'X^a' = 'EXP(LN(X)) ^ a <1>
= 'EXP(LN(X)*a)' <2: (A^B) ^ C = A ^ (B*C)>
= 'EXP(a*LN(X))' <3>

But now substitute X by 0 in <3> :
We get 'EXP(a*LN(0))' <4>

But in <4> LN(0)= -infinity
So that <4> becomes EXP(0 * -infinity) <5>
And in <5> (0 * -infinity) gives '?'
And of course EXP(0 * -infinity) gives '?'

Therefore, considering all possible cases,
<3> cannot be simplified further in the HP50G logic.

Now, considering the above with all possible cases, with
X=EXP(LN(X) <0>
X^a = EXP(a*LN(X)) <3>

then X * X^a <6>
=EXP(LN(X) * EXP(a*LN(X)) <7>
And <7> simplified (by SIMPLIFY)
gives X * EXP(a*LN(X)) <8>

And expression <8> EXPAND
gives back X * X^a <6> (with no further simplification, as already mentioned).

B)
On the other hand, let's check, the other way, if:
X*X^a =? X^(a+1) <9>

Let X=a=0 in above <9>
We get 0*(0^0) =? 0^1 = 0.
<=> 0 * {1 or 0, no other possible output claim like -infinity or +infinity } =? 0
<=> 0 * "any number" =? 0
<=> 0 = 0

And here, by my development,
X*X^a =? X^(a+1) <9>
becomes, also for X=a=0:
X*X^a = X^(a+1) <13>

C)
Let now X=0, a=-1 in <13>.
<13> will become
0*0^(-1) =? 0^(-1+1)
<=> 0*1/0 =? 0^0
<=> 0/0 =? 0^0 <14>
<=>0/0 =? {1, 0} <15>
<=> '?' =? {1 or 0} <16>
And '?' cannot be said to be equal to {1 or 0} <17>

By this way of reasoning we cannot
conclude in <14/15/16> with an equality sign (= instead of =? >between left and right result in <14/15/16>.

Therefore, all in all, HP50G plays — correctly — safety, though it might appear unnesseary (or excessively careful) in all but one case (the exception being when X=a=0).

PS

If X^a= 1 when X=a=0 (0^0=1) <101>
then indefinite integral of '0^0' = <102>
indefinite integral of '1' = X... <103>
But X=0, then indefinite integral of '1'=0 <104>
But derivative of any constant = 0.
Therefore, derivative of <104> = derivative of 0=0 <105>
but that latter result (0) ≠ back the special function to be integrated '0^0' =1 (according to 101>
—> 0^0=1 gives a contradiction for the above given special integral

Instead if 0^0=0 <106>
then indefinite integral of 'X^a' with X=a=0
= undefinite integral of 0^0
= undefinite integral of '0'
= any constant.
And derivative of any constant =0... =0^0=0
—>0^0=0 gives no contradiction for the above given indefinite special integral.