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

[Gil]

On exact mode
X^a'
INTVX
and we get '1/(a+1)*X*X^a'

What command should I use afterwards to simplify the expression and collect the two powers of x (1 and a into a single power =1+a)?

Thanks for your help.

Regards,
Gil

[gor1060]

Hello!
   
   

[peacecalc]

Hello gor1060,

I'm very fond of the MATCH(arrow up or down) command, the user has the possibility to add his own cas commands operating on literals. Even the hp prime has no such an excellent tool. I'm shure that someone can write a program in PPL, that can do the same, but it is much more complicated, I think.

Edit was necessary, because I forgot a word.

[Gil]

Sorry, but I could not enter the two expressions in {}.

How did you enter them?

I did not put both expressions with '... '..
Now it's OK.

My wish was not to teach the calculator to simplify, but rather ask the calculator to simplify the found expression by a specific command.

[peacecalc]

Hallo Gil,
I failed with the inserted cas commands to solve the problem, too. The HP 50g is a limited machine. It is a pity HP 50g fails in such an easy case.
But put the command from gor1060 in a program and you have in future an extension of cas. It is not nice, but it helps.

[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.

[Albert Chan]

(12-02-2023 02:21 PM)Gil Wrote:  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).

By this logic, HP50G should not be able to simplify(x^2/x) = x (but it does!)
I think edge case should be treated as limit, a=0, x→0

limit(x*x^0, x=0) = limit(x*1, x=0) = 0
limit(x^(0+1), x=0) = limit(x^1, x=0) = 0

What if a=-1, exactly?

limit(x*x^(-1), x=0) = limit(x/x, x=0) = 1
limit(x^(-1+1), x=0) = limit(x^0, x=0) = 1

--> x*x^a == x^(a+1), even edge cases.

[Gil]

How can we not agree with Albert Chan?
Short, clear and convincing.