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Re: Coaxing N[] to work
*To*: mathgroup at smc.vnet.net
*Subject*: [mg73394] Re: Coaxing N[] to work
*From*: Scott Hemphill <hemphill at hemphills.net>
*Date*: Thu, 15 Feb 2007 05:01:40 -0500 (EST)
*References*: <equo6r$in0$1@smc.vnet.net>
*Reply-to*: hemphill at alumni.caltech.edu
<p at dirac.org> writes:
> Sometimes N[,] doesn't appear to work. Like here:
>
>
> x = {2.0, 3.0, 5.0};
> A = { {6.0, 2.0, 1.0}, {2.0, 3.0, 1.0}, {1.0, 1.0, 1.0} };
> For[ k=0, k<15, ++k,
> lambda = x.A.x/(x.x);
> y = LinearSolve[A,x];
> x = y / Norm[y,Infinity];
> ]
> N[lambda, 30]
>
>
> The output is:
>
> Out[5]= 0.578933
>
> I was expecting 30 digits. Why did N[] ignore my request for 30 digits?
Because lambda had already been calculated, and it didn't have 30 digits.
See how Mathematica prints the following numbers:
In[1]:= 2
Out[1]= 2
In[2]:= 2.0
Out[2]= 2.
In[3]:= 2`30
Out[3]= 2.00000000000000000000000000000
Now see what the precision of each number is:
In[4]:= Precision[2]
Out[4]= Infinity
In[5]:= Precision[2.0]
Out[5]= MachinePrecision
In[6]:= Precision[2`30]
Out[6]= 30.
Here's your problem, redone with infinite precision:
In[7]:= x = {2, 3, 5};
In[8]:= A = { {6, 2, 1}, {2, 3, 1}, {1, 1, 1} };
In[9]:= For[ k=0, k<15, ++k,
lambda = x.A.x/(x.x);
y = LinearSolve[A,x];
x = y / Norm[y,Infinity];
]
In[10]:= lambda
1102158619423970036337
Out[10]= ----------------------
1903774504398915184457
In[11]:= N[lambda, 30]
Out[11]= 0.578933385691052787623495851172
Here's your problem, redone with specified precision. Whenever a number
with infinite precision is arithmetically combined with a number with
finite precision, the result has finite precision. If I had wanted, I
could have set all the elements in x and A to finite precision by using
SetPrecision, e.g., "x = SetPrecision[ {2, 3, 5}, 30 ];".
In[12]:= x = {2`30, 3, 5};
In[13]:= A = { {6, 2, 1}, {2, 3, 1}, {1, 1, 1} };
In[14]:= For[ k=0, k<15, ++k,
lambda = x.A.x/(x.x);
y = LinearSolve[A,x];
x = y / Norm[y,Infinity];
]
In[15]:= lambda
Out[15]= 0.5789333856910527876234959
In[16]:= Precision[%]
Out[16]= 24.9696
Note that some precision was lost due to round off error. You can't
get more precision after the number has already been calculated:
In[17]:= N[%15,100]
Out[17]= 0.5789333856910527876234959
Scott
--
Scott Hemphill hemphill at alumni.caltech.edu
"This isn't flying. This is falling, with style." -- Buzz Lightyear
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