I have to import a large integer (1000 digits) into matlab to run some calculations on it. However, when I import it I seem to loose accuracy due to the fact that matlab uses the scientific notation.
Is there any way that I can get the actual integer?
Here's the actual data I have to import:
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Such a large integer cannot be represented in IEEE floating point standard. Check out this answer for the largest double that can be represented without losing precision (its 1.7977e+308). That can be obtained by typing realmax in MATLAB.
You can use vpi (available here, as mentioned in comment) or you can use the MATLAB in-built vpa.
This is how you use vpa
R=vpa('7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450');
You can check the following:
vpa('R+1000-R')
The answer of the above is 1000 as expected. Do not forget to put your expression in quotes. Otherwise, you are passing inifinity to vpa instead of the 1000 digit number.
If you want to use vpi, its a beautiful toolbox, go ahead, download it. Go into its root directory and run the following command:
a=vpi('7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450')
Well, the advantage with vpi is as follows:
The output of vpi:
a=vpi(<<Your 1000 digit number in quotes>>); %output prints 1000 digits on screen.
The output of vpa:
R=vpa(<<Your 1000 digit number in quotes>>);
this prints:
R =
7.3167176531330624919225119674427e999
Also, with vpi, you can do something like this:
a=vpi('7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450')
b=a+1
b-a %output of this yields 1.
I somehow cannot do the operation of b-a in vpa and obtain the answer 1.
Related
I am converting a program from MATLAB 2012 to 2016. I've been getting some strange errors, which I believe some of are due to a lack of precision in MATLAB functions.
For instance, I have a timeseries oldTs as such:
Time Data
-----------------------------
1.00000000000000001 1.277032377439511
1.00000000000000002 1.277032378456123
1.00000000000000003 1.277032380112478
I have another timeseries newTs with similar data, but many more rows. oldTs may have half a million rows, whereas newTs could have a million. I want to interpolate the data from the old timeseries with the new timeseries, for example:
interpolatedTs = interp(oldTs.time, oldTs.data, newTs.time)
This is giving me an error: x values must be distinct
The thing is, my x values are distinct. I think that MATLAB may be truncating some of the data, and therefore believing that some of the data is not unique. I found that other MATLAB functions do this:
test = [1.00000000000000001, 1.00000000000000002, 1.0000000000000000003]
unique(test)
ans =
1
test2 = [10000000000000000001, 10000000000000000002, 10000000000000000003]
unique(test2)
ans =
1.000000000000000e+19
MATLAB thinks that this vector only has one unique value in it instead of three! This is a huge issue for me, as I need to maintain the highest level of accuracy and precision with my data, and I cannot sacrifice any of that precision. Speed/Storage is not a factor.
Do certain MATLAB functions, by default, truncate data at a certain nth decimal? Has this changed from MATLAB 2012 to MATLAB 2016? Is there a way to force MATLAB to use a certain precision for a program? Why does MATLAB do this to begin with?
Any light shed on this topic is much appreciated. Thanks.
No, this has not changed since 2012, nor since the very first version of MATLAB. MATLAB uses, and has always used, double precision floating point values by default (8 bytes). The first value larger than 1 that can be represented is 1 + eps(1), with eps(1) = 2.2204e-16. Basically you have less than 16 decimal digits to play with. Your value 1.00000000000000001 is identical to 1 in double precision floating point representation.
Note that this is not something specific to MATLAB, it is a standard that your hardware conforms to. MATLAB simply uses your hardware's capabilities.
Use the variable precision arithmetic from the Symbolic Math Toolbox to work with higher precision numbers:
data = [vpa(1) + 0.00000000000000001
vpa(1) + 0.00000000000000002
vpa(1) + 0.00000000000000003]
data =
1.00000000000000001
1.00000000000000002
1.00000000000000003
Note that vpa(1.00000000000000001) will not work, as the number is first interpreted as a double-precision float value, and only after converted to VPA, but the damage has already been done at that point.
Note also that arithmetic with VPA is a lot slower, and some operations might not be possible at all.
I am trying to calculate some integrals that use very high power exponents. An example equation is:
(-exp(-(x+sqrt(p)).^2)+exp(-(x-sqrt(p)).^2)).^2 ...
./( exp(-(x+sqrt(p)).^2)+exp(-(x-sqrt(p)).^2)) ...
/ (2*sqrt(pi))
where p is constant (1000 being a typical value), and I need the integral for x=[-inf,inf]. If I use the integral function for numeric integration I get NaN as a result. I can avoid that if I set the limits of the integration to something like [-20,20] and a low p (<100), but ideally I need the full range.
I have also tried setting syms x and using int and vpa, but in this case vpa returns:
1.0 - 1.0*numeric::int((1125899906842624*(exp(-(x - 10*10^(1/2))^2) - exp(-(x + 10*10^(1/2))^2))^2)/(3991211251234741*(exp(-(x - 10*10^(1/2))^2) + exp(-(x + 10*10^(1/2))^2)))
without calculating a value. Again, if I set the limits of the integration to lower values I do get a result (also for low p), but I know that the result that I get is wrong – e.g., if x=[-100,100] and p=1000, the result is >1, which should be wrong as the equation should be asymptotic to 1 (or alternatively the codomain should be [0,1) ).
Am I doing something wrong with vpa or is there another way to calculate high precision values for my integrals?
First, you're doing something that makes solving symbolic problems more difficult and less accurate. The variable pi is a floating-point value, not an exact symbolic representation of the fundamental constant. In Matlab symbolic math code, you should always use sym('pi'). You should do the same for any other special numeric values, e.g., sqrt(sym('2')) and exp(sym('1')), you use or they will get converted to an approximate rational fraction by default (the source of strange large number you see in the code in your question). For further details, I recommend that you read through the documentation for the sym function.
Applying the above, here's a runnable example:
syms x;
p = 1000;
f = (-exp(-(x+sqrt(p)).^2)+exp(-(x-sqrt(p)).^2)).^2./(exp(-(x+sqrt(p)).^2)...
+exp(-(x-sqrt(p)).^2))/(2*sqrt(sym('pi')));
Now vpa(int(f,x,-100,100)) and vpa(int(f,x,-1e3,1e3)) return exactly 1.0 (to 32 digits of precision, see below).
Unfortunately, vpa(int(f,x,-Inf,Inf)), does not return an answer, but a call to the underlying MuPAD function numeric::int. As I explain in this answer, this is what can happen when int cannot obtain a result. Normally, it should try to evaluate the the integral numerically, but your function appears to be ill-defined at ±∞, resulting in divide by zero issues that the variable precision quadrature methods can't handle well. You can evaluate the integral at wider bounds by increasing the variable precision using the digits function (just remember to set digits back to the default of 32 when done). Setting digits(128) allowed me to evaluate vpa(int(f,x,-1e4,1e4)). You can also more efficiently evaluate your integral over a wider range via 2*vpa(int(f,x,0,1e4)) at lower effective digits settings.
If your goal is to see exactly how much less than one p = 1000 corresponds to, you can use something like vpa(1-2*int(f,x,0,1e4)). At digits(128), this returns
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000086457415971094118490438229708839420392402555445545519907545198837816908450303280444030703989603548138797600750757834260181259102
Applying double to this shows that it is approximately 8.6e-89.
i have code and use double function several time to convert sym to double.to increase precision , I want to use digits function.
I want to know it is enough that I write digits in the top of code or I must write digits in above of every double function.
digits set's the precision until it is changed again. Calling digits() without any input you get the precision to verify it's set correct.
In many cases digis has absoluetly no influence on symbolic variables because an analytical solution is found. This means there are no precision errors unless you convert to double. When convertig, digits should be set to at least 16 because this matches double precision.
as i know to get zero mean vector from given vector,we should substract mean of given vector from each memeber of this vector.for example let us see following example
r=rand(1,6)
we get
0.8687 0.0844 0.3998 0.2599 0.8001 0.4314
let us create another vector s by following operation
s=r-mean(r(:));
after this we get
0.3947 -0.3896 -0.0743 -0.2142 0.3260 -0.0426
if we calculate mean of s by following formula
mean(s)
we get
ans =
-5.5511e-017
actually as i have checked this number is very small
-5.5511*exp(-017)
ans =
-2.2981e-007
so we should think that our vector has mean zero?so it means that that small deviation from 0 is because of round off error?for exmaple when we are creating white noise or such kind off random uncorrelated sequence of data,actually it is already supposed that even for such small data far from 0,it has zero mean and it is supposed in this case that for example for this case
-5.5511e-017 =0 ?
approximately of course
e-017 means 10 to the power of -17 (10^-17) but still the number is very small and hypothetically it is 0. And if you type
format long;
you will see the real precision used by Matlab
Actually you can refer to the eps command. Although matlab uses double that can encode numbers down to 2.2251e-308 the precission is determined size of the number.
Use it in the format eps(number) - it tell you the how large is the influence of the least significant bit.
on my machine eg. eps(0.3) returns 5.5511e-17 - exactly the number you report.
I wanted to create a vector with three values 1/6, 2/3 and 1/6. Obviously I Matlab has to convert these rational numbers into real numbers but I expected that it would maximize the precision available.
It's storing the values as doubles but it's storing them as -
b =
0.1667 0.6667 0.1667
This is a huge loss of precision. Isn't double supposed to mean 52 bits of accuracy for the fractional part of the number, why are the numbers truncated so severly?
The numbers are only displayed that way. Internally, they use full precision. You can use the format command to change display precision. For example:
format long
will display them as:
0.166666666666667 0.666666666666667 0.166666666666667
So the answer is simple; there is no loss of precision. It's only a display issue.
You can read the documentation on what other formats you can use to display numbers.
you can not store values as 1/2 or 1/4 or 1/6 in to a Double variable... these are stored as decimals behind the system; if you want to store these values , try storing it as string that would work;
Whenever you want to make mathematical calculation using these strings then convert the value into number and continue....