On 2020-05-16 14:48, William Michels via perl6-users wrote:
Yes, ** stands for exponentiation
Thank you!
Hi Todd,
Yes, ** stands for exponentiation. And exponentiation has higher
precedence than multiplication. See below (I dropped a few zeros to
help clarify):
> put 2 * 10 ** 20
2
> put (2 * 10) ** 20
1048576
> put 2 * (10 ** 20)
2
> 20
On 2020-05-14 22:26, Peter Pentchev wrote:
And what is `2 * 100 ** 2000 `? Is that `(2 x 100)^ 2000`
((2 times 100) to the 2000 power?
Point 1: exponentiation has a higher priority than multiplication.
Point 2:https://rosettacode.org/wiki/Integer_roots
Any chance of you answering the
On 2020-05-15 12:09, Bruce Gray wrote:
On May 14, 2020, at 4:36 PM, ToddAndMargo via perl6-users
wrote:
On 2020-05-14 08:13, Bruce Gray wrote:
On May 14, 2020, at 7:27 AM, ToddAndMargo via perl6-users
wrote:
Hi All,
1) how do I get 40 or more digits out of sqrt?
—snip—
Use an Integer
> On May 14, 2020, at 4:36 PM, ToddAndMargo via perl6-users
> wrote:
>
> On 2020-05-14 08:13, Bruce Gray wrote:
>>> On May 14, 2020, at 7:27 AM, ToddAndMargo via perl6-users
>>> wrote:
>>>
>>> Hi All,
>>>
>>> 1) how do I get 40 or more digits out of sqrt?
>> —snip—
>> Use an Integer Root
On Thu, May 14, 2020 at 02:36:30PM -0700, ToddAndMargo via perl6-users wrote:
> On 2020-05-14 08:13, Bruce Gray wrote:
> >
> >
> > > On May 14, 2020, at 7:27 AM, ToddAndMargo via perl6-users
> > > wrote:
> > >
> > > Hi All,
> > >
> > > 1) how do I get 40 or more digits out of sqrt?
> >
> >
On 2020-05-14 05:51, Tobias Boege wrote:
On Thu, 14 May 2020, ToddAndMargo via perl6-users wrote:
Hi All,
1) how do I get 40 or more digits out of sqrt?
Meaningful digits? Not possible as sqrt uses limited precision. I think
the IEEE 754 doubles that I would suspect to be used internally
On 2020-05-14 08:13, Bruce Gray wrote:
On May 14, 2020, at 7:27 AM, ToddAndMargo via perl6-users
wrote:
Hi All,
1) how do I get 40 or more digits out of sqrt?
—snip—
Use an Integer Root algorithm on ($number-you-want-the-root-of * 100 **
$number-of-digits-you-want), then shift the
> On May 14, 2020, at 7:27 AM, ToddAndMargo via perl6-users
> wrote:
>
> Hi All,
>
> 1) how do I get 40 or more digits out of sqrt?
—snip—
Use an Integer Root algorithm on ($number-you-want-the-root-of * 100 **
$number-of-digits-you-want), then shift the decimal point of the result.
On 2020-05-14 05:51, Tobias Boege wrote:
On Thu, 14 May 2020, ToddAndMargo via perl6-users wrote:
Hi All,
1) how do I get 40 or more digits out of sqrt?
Meaningful digits?
In this instance, I do not care about meaningful. I only
care about the noise. Just has to be repeatable.
I may
On Thu, 14 May 2020, ToddAndMargo via perl6-users wrote:
> Hi All,
>
> 1) how do I get 40 or more digits out of sqrt?
>
Meaningful digits? Not possible as sqrt uses limited precision. I think
the IEEE 754 doubles that I would suspect to be used internally are capped
way below 40 significant
Hi All,
1) how do I get 40 or more digits out of sqrt?
2) how to I assist those bytes to a 40 (or more)
byte long Buf?
This obviously does not work:
my Num $x = 3.sqrt; my Buf $y = Buf.new($x)
Type check failed in initializing element #0 to Buf; expected uint8 but
got Num
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