Read the e-notation part.

RAND produces a UNIFORMLY distributed random number. Do you know what the probability that a uniformly distributed random number, when selected from the interval [1e-40,1e-6], will actually lie in the sub-interval [1e-40,1e-39]? Roughly... (1e-39 - 1e-40)/(1e-6 - 1e-40) ans = 9e-34 I said roughly because MATLAB cannot actually compute the difference (1e-6-1e-40) completely accurately in double precision, but to compute the number exactly would require more effort than it is worth given the tiny difference. So seriously, what would you expect? If you were to compute many billions of such numbers, you would still expect never to see such a result. Worse, it would never actually happen anyway, since that range of numbers is wider than the range you can compute with doubles. Instead, you might decide to compute numbers randomly and uniformly in the interval [-40,-6]. Then raise 10 to that power. The result will NOT be uniform of course. It will have the properties that you seem to want however.

Hi Christian, Try this: title(sprintf('Surface Ions per nm^2: %0.2f x 10^%i', 10^mod(log10(surfaceIons),1),floor(log10(surfaceIons))))

You may wish to give the command format short g or format long g

"format short" or "format long". See also your Preferences -> Command Window Preferences, as the default can be set there.

with format long;x=1:10 = 1 2 3 4 5 6 7 8 9 10 vpa only adds one zero: y=vpa(x,5) = [ 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0] y=vpa(x,10) = [ 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0] floating point boundaries (without vpa): flintmax = 9.007199254740992e+15 intmax = 2147483647 intmin = -2147483648 with fixed point precision objects, details here: you get more than 4 zeros: a = fi(pi, 1, 8, 3) a = 3.125000000000000 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 3 but only if it is rounding non-zero decimals: a = fi(1, 1, 8, 3) = 1 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 3 a = fi(1.0000, 1, 8, 3) = 1 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 3 tried 1.0001 and then -0001 but did not work: b = fi(1.0001, 1) b = 1.000122070312500 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 14 b-.0001 Warning: The behavior of adding or subtracting a fi and a non-fi operand has changed. The non-fi operand is now cast to the data type of the fi operand. See the release notes for information on retaining the old behavior of best precision scaling cast. In - (line 37) = 1 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 17 FractionLength: 14 c=b-.0001 = 1 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 17 FractionLength: 14 b2=fi(0.0001,1) = 1.000016927719116e-04 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 28 b-b2 = 1.000022068619728 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 31 FractionLength: 28 d=b-b2 = 1.000022068619728 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 31 FractionLength: 28 d2=vpa(d,4) = 1.0 If you really want to present the vector exactly with 4 zero decimals, you can translate each figure to char, x=[1:1:4] x_char=char(x+48)' = 1 2 3 4 and then add '.000' x2=repmat(['.0000'],4,1) = .0000 .0000 .0000 .0000 x_char=[x_char x2] = 1.0000 2.0000 3.0000 4.0000 If you find this answer of any help solving your question, please click on the thumbs-up vote link, thanks in advance John jgb2012@sky.com

reshape( 1:100, 10, 10) .'

A=[1;2;3] B=A'