axiom-developer
[Top][All Lists]
Advanced

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

## Re: [Axiom-developer] Schaums help

 From: Doug Stewart Subject: Re: [Axiom-developer] Schaums help Date: Sat, 03 May 2008 10:05:49 -0400 User-agent: Thunderbird 2.0.0.12 (X11/20080227)

```Doug Stewart wrote:
```
```root wrote:
```
```My copy of Schaums (1968, printing 4) shows

14:334:

int(1/(x*sqrt(x^n-a^n)),x) == 2/(n*sqrt(a^n))*acos(sqrt(a^n/x^n))

It seems this cannot be the answers.
Can someone with a later version please check for a typo?

Tim

_______________________________________________
Axiom-developer mailing list
address@hidden
http://lists.nongnu.org/mailman/listinfo/axiom-developer

```
```My schaums shows that answer.
also usind Maxima to do the derivative  I get the LHS.
(%i5) diff(2/(n*sqrt(a^n))*acos(sqrt(a^n/x^n)),x);
(%o5) (a^n*x^(-n-1))/(sqrt(a^n)*sqrt(a^n/x^n)*sqrt(1-a^n/x^n))
(%i6) radcan(%);
(%o6) 1/(x*sqrt(x^n-a^n))
```
```
If you compute
aa:=integrate(1/(x*sqrt(x^n-a^n)),x)
bb:=2/(n*sqrt(a^n))*acos(sqrt(a^n/x^n))
cc1:=aa.1-bb
cc2:=aa.2-bb

```
Can you find a simplification path (in Axiom) such that either cc1 or cc2 simplify to a constant?
```
Alternatively, can you use Maxima to find the constant?

I'm failing to do either, although I'm still trying.

Tim

```
```
Maxima seems to give a wrong answer for this integration.

(%i1) aa:integrate(1/(x*sqrt(x^n-a^n)),x);
Is  a  positive or negative?p;

(%o1) (2*atan(sqrt(x^n-a^n)/a^(n/2)))/(a^(n/2)*n)

and
(%i2) aa:integrate(1/(x*sqrt(x^n-a^n)),x);
Is  a  positive or negative?n;
```
(%o2) log((2*sqrt(x^n-a^n)-2*sqrt(-a^n))/(2*sqrt(x^n-a^n)+2*sqrt(-a^n)))/(sqrt(-a^n)*n)
```

I'm not good enough at simplifying to help you.
Doug

_______________________________________________
```
```with help from Maxima
```
```>Can you show me how to start with
>
>(2*atan(sqrt(x^n-a^n)/a^(n/2)))/(a^(n/2)*n)
>
>and differentiate it to get
>
>1/(x*sqrt(x^n-a^n)
>
>using Maxima.
```
```
(%i4) (2*atan(sqrt(x^n-a^n)/a^(n/2)))/(a^(n/2)*n)\$
(%i5) ratsimp(diff(%,x));
(%o5) 1/(x*sqrt(x^n-a^n))

Barton

so Maxima is not wrong.

Doug

```

reply via email to

 [Prev in Thread] Current Thread [Next in Thread]