how do i solve power(5,1.3)?
def power(nbr, po):
if po==0:
return 1
if po>0:
return nbr*power(nbr, po-1)
if po<0:
return 1/power(nbr, -1*po)
also i found a link which states 0^0 isnt 1 even though every
calculator ive tried says it is.
it doesnt say what it is but i presume 0 then.
but it seems the dude is wrong and it is 1?
dont run the code with decimals, it will never leave the function, u
have to restart the shell(if using the standard python ide)
8  3104 
On Apr 16, 4:19*pm, skanem...@yahoo .se wrote:
how do i solve power(5,1.3)?
[...]
>
also i found a link which states 0^0 isnt 1 even though every
calculator ive tried says it is.
it doesnt say what it is but i presume 0 then.
but it seems the dude is wrong and it is 1?
>>5**1.3
8.1032829834638 136
>>0**0
1
>>0.**0.
1.0
>>from decimal import Decimal
Decimal(0)**D ecimal(0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/System/Library/Frameworks/Python.framewor k/Versions/2.5/lib/
python2.5/decimal.py", line 1755, in __pow__
return context._raise_ error(InvalidOp eration, '0 ** 0')
File "/System/Library/Frameworks/Python.framewor k/Versions/2.5/lib/
python2.5/decimal.py", line 2325, in _raise_error
raise error, explanation
decimal.Invalid Operation: 0 ** 0
Most mathematicians consider 0**0 to be either 1 or undefined. Which
answer you get depends on who you ask (or in Python, whether you're
working with floats or Decimals, as you see above).
Mark sk*******@yahoo .se wrote:
how do i solve power(5,1.3)?
Is this a trick question? OK, I'll bite:
>>5 ** 1.3
8.1032829834638 136
>>>
>
def power(nbr, po):
if po==0:
return 1
if po>0:
return nbr*power(nbr, po-1)
if po<0:
return 1/power(nbr, -1*po)
also i found a link which states 0^0 isnt 1 even though every
calculator ive tried says it is.
it doesnt say what it is but i presume 0 then.
but it seems the dude is wrong and it is 1?
Of the possible results of 0 ** 0 (i.e. 1, 0, and NaN), 1 seems to be
the least implausible. It allows X ** 0 to be 1 for all X.
>
dont run the code with decimals, it will never leave the function, u
have to restart the shell(if using the standard python ide)
I presume that by "decimals", you mean "numbers that are not integers".
So you've got it into an infinite loop. Have you tried tracing through
the first 5 or 6 gyrations? This can be done by
(a) putting a debugger breakpoint just after the start of the function
(b) using something like:
print "power(%.6f , %.6f)" % (nbr, po))
junk = raw_input("Pres s <Enterto continue -")
(c) using a pencil and a piece of scrap paper, write down what is
happening as a typical function call is executed e.g.
power(5, 1.3) =5 * power(5, 0.3)
power(5, 0.3) =5 * power(5, -0.7)
power(5, -0.7) =1 / power (5, 0.7)
etc
Then work out what extra condition you would have to test to stop it
doing that. Then work out how to calculate the return value when that
condition is true.
HTH,
John
En Wed, 16 Apr 2008 19:21:18 -0300, John Machin <sj******@lexic on.net>
escribió:
sk*******@yahoo .se wrote:
>also i found a link which states 0^0 isnt 1 even though every
calculator ive tried says it is.
it doesnt say what it is but i presume 0 then.
but it seems the dude is wrong and it is 1?
Of the possible results of 0 ** 0 (i.e. 1, 0, and NaN), 1 seems to be
the least implausible. It allows X ** 0 to be 1 for all X.
But a result of 0 would allow 0 ** X to be 0 for all X. (btw, this is the
reason lim(x**x) for x->0 does not exist)
--
Gabriel Genellina
On Apr 16, 5:49*pm, "Gabriel Genellina" <gagsl-...@yahoo.com.a r>
wrote:
En Wed, 16 Apr 2008 19:21:18 -0300, John Machin <sjmac...@lexic on.net*
escribió:
skanem...@yahoo .se wrote:
also i found a link which states 0^0 isnt 1 even though every
calculator ive tried says it is.
it doesnt say what it is but i presume 0 then.
but it seems the dude is wrong and it is 1?
Of the possible results of 0 ** 0 (i.e. 1, 0, and NaN), 1 seems to be
the least implausible. It allows X ** 0 to be 1 for all X.
But a result of 0 would allow 0 ** X to be 0 for all X. (btw, this is the *
reason lim(x**x) for x->0 does not exist)
Where this has come up in my research, X**0 being a multiplicative
identity is far more important than 0**X being an additive identity.
>
--
Gabriel Genellina
"Mark Dickinson" <di******@gmail .comwrote in message
news:fc******** *************** ***********@y18 g2000pre.google groups.com...
On Apr 16, 4:19 pm, skanem...@yahoo .se wrote:
how do i solve power(5,1.3)?
[...]
>
also i found a link which states 0^0 isnt 1 even though every
calculator ive tried says it is.
it doesnt say what it is but i presume 0 then.
but it seems the dude is wrong and it is 1?
Define a**b as 1 multiplied by a b times. Then a**0 is clearly 1,
regardless of a.
But some do disagree.
| decimal.Invalid Operation: 0 ** 0
I would think of this as a bug unless the standard Decimal follows demands
this.
tjr
actually that 0**0 statement was wrong. 0**0 = 1 and should be.
On 17 avr, 00:49, "Gabriel Genellina" <gagsl-...@yahoo.com.a rwrote:
En Wed, 16 Apr 2008 19:21:18 -0300, John Machin <sjmac...@lexic on.net>
escribió:
skanem...@yahoo .se wrote:
also i found a link which states 0^0 isnt 1 even though every
calculator ive tried says it is.
it doesnt say what it is but i presume 0 then.
but it seems the dude is wrong and it is 1?
Of the possible results of 0 ** 0 (i.e. 1, 0, and NaN), 1 seems to be
the least implausible. It allows X ** 0 to be 1 for all X.
But a result of 0 would allow 0 ** X to be 0 for all X. (btw, this is the
reason lim(x**x) for x->0 does not exist)
lim(x**x) for x->0+ is well defined, exists and equals 1. [1]
As x**x is continuous in 0+, it is widely customary to have: 0**0:=1
[1] Recall that x**x := exp(x*log(x))
The limit of x*log(x) for x->0 is 0 [2] therefore lim(x**x) for x->0
is 1.
[2] Let y = 1/x; x*log(x)= -log(y)/y and the limit of log(y)/y for y->
+inf is 0.
On Apr 16, 11:03*pm, "Terry Reedy" <tjre...@udel.e duwrote:
| decimal.Invalid Operation: 0 ** 0
I would think of this as a bug unless the standard Decimal follows demands
this.
It does. From http://www2.hursley.ibm.com/decimal/daops.html#refpower
:
"If both operands are zero, or if the left-hand operand is less than
zero and the right-hand operand does not have an integral value[7] or
is infinite, an Invalid operation condition is raised, the result is
[0,qNaN], and the following rules do not apply."
I'm hoping that this will change with the next update of the standard.
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