Matrix X.^23 without approximation

I want to get the answer with no approximation,

I have X=[25 24 23;27 8 11]

I want X.^23 without approximation,help me please.

Answer 1

[Torsten]

Apply Titus' answer under

http://www.mathworks.com/matlabcentral/answers/227104-25-23-without-approximation

to each element of X.

Best wishes

Torsten.

[Sakunrat Jaejaima]

Same applie ?

[Torsten]

??

[Titus Edelhofer]

X.^23

is the same as

[x(1,1)^23  x(1,2)^23
 x(2,1)^23 ...]

so apply the technique in my answer to the last time you asked this question to each single entry in your matrix.

Answer 2

[bio lim]

What do you mean by without approximation? Are you talking about scientific notations? If so, you can change the format using, format style.

X=[25 24 23;27 8 11];
format bank % I am guessing this is the format you want.
X.^23

[Sakunrat Jaejaima]

Ex. 25^23 without approximation .I want to be integer

[bio lim]

X=[25 24 23;27 8 11];
% The longest digit
digit_number = floor(log10(X(2,1)^23));
output = var((X.^23), digit_number))

I don't have access to matlab right now, so check and comment if there is an error.

[Sakunrat Jaejaima]

it error .

[bio lim]

output = vpa((X.^23),digit_number);

Answer 3

[Steven Lord]

You will need to compute in higher precision than double. 25^23 is large enough that not all integers that are that large can be exactly represented.

 25^23 > flintmax

In fact, 25^23 is so large that even 1,000,000 is negligible compared to it.

 z1 = 25^23;
 z2 = z1 + 1000000;
 z1 == z2 % will return TRUE, this is NOT a bug!

Think of it as though you gave Bill Gates a $5 bill. He's so rich ($78.8 billion according to Wikipedia) that $5 doesn't change his net worth at all. [Contacting the necessary accountants to change his net worth would cost him more than the $5 you gave him!]

You can compute symbolically:

 X = sym([25 24 23;27 8 11]);
 X.^23

Or if you're doing this as part of computing 25^23 mod N for some value N, don't compute 25^23 first then compute MOD. That's the straightforward approach described on the "Modular exponentiation" Wikipedia page, but it breaks down when the quantity whose modulus you're taking gets too large. Instead apply one of the other techniques described on that page. For small exponents, the memory efficient method is easy to write; the right-to-left is a little more difficult, but using BITGET you can do it reasonably easily in MATLAB.

[Sakunrat Jaejaima]

Thank you so mush.