Not a silly question at all.
No. Every verb has access to its own private namespace and to the
public namespaces. No explicit verb can access the private namespace of
another verb. Where the verb is defined makes no difference.
If you want v to access nouns from u, you must pass them as ar
I have an explicitly defined verb v say
Verb v uses, inside its definition, another verb u say
Verb u depends on a constants which are calculated inside v.
Can I “redefine” u inside v?
Apologies if this is an obviously silly question.
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Also if you multiply by x to get
2 + 4.9x = 0
You can solve using p.
p. 2 4.9
┌───┬─┐
│4.9│_0.408163│
└───┴─┘
On Sun, 25 Oct 2020, 09:47 Martin Kreuzer, wrote:
> N is at most times sensitive to its start value ...
> it does work in this case (but as you stated, it will no
N is at most times sensitive to its start value ...
it does work in this case (but as you stated, it will not be
neccessary in this case):
(4.9+2&%) N^:1 (_0.1)
_0.1755
(4.9+2&%) N^:2 (_0.1)
_0.275539
(4.9+2&%) N^:4 (_0.1)
_0.403614
(4.9+2&%) N^:14 (_0.1)
_0.408163
(4.9+2&%) N^:_
Jd release 4.36 available:
https://code.jsoftware.com/wiki/Jd/Release
Read statements with only a single table and an empty where run faster. In
particular, this lets aggregations with a by clause that apply to an entire
table run 3 times faster.
---
I just realized that I don't need Newton:
2/x + 4.9 = 0?
2/x = 49/10
2 = 49/10 * x
x = 2/(49/10) = 2*(10/49) = 20/49 = 0.4081632653
Skip Cave
Cave Consulting LLC
On Sat, Oct 24, 2020 at 10:12 AM Skip Cave wrote:
> The Newton Raphson root finder using the new J definition of derivative is:
>
The Newton Raphson root finder using the new J definition of derivative is:
N=: 1 : '- u % u deriv_jcalculus_ 1'
How would I use N to solve for x in the equation 2/x + 4.9 = 0?
Skip Cave
Cave Consulting LLC
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