I have defined a really basic function in matlab. It takes no input and returns an array of 10 floating point numbers.
The problem I have is that when I run the function to return the array I want I get incorrect values, however when I substitute in a value and simply print out the value from within the function I get the correct answer?!
I've posted samples from the code below:
% Calculate the terms in our expression
FirstTerm = sin(Alpha)*(atan(x+d)-atan(x-d));
SecondTerm = cos(Alpha)*0.5*log(((x+d).^2+h.^2)/((x-d).^2+h.^2));
% Combine and return result
Result = 2 * (FirstTerm - SecondTerm)
FirstTermTemp = sin(Alpha)*(atan(-8+d)-atan(-8-d));
SecondTermTemp = cos(Alpha)*0.5*log(((-8+d).^2+h.^2)/((-8-d).^2+h.^2));
ResultTemp = 2 * (FirstTermTemp - SecondTermTemp)
The array I want to calculate for starts at -8 so the results should match. Does anyone have any idea why they wouldn't?
Cheers
Jack
You have left off a . before your /
% //Calculate the terms in our expression
FirstTerm = sin(Alpha)*(atan(x+d)-atan(x-d));
SecondTerm = cos(Alpha)*0.5*log(((x+d).^2+h.^2)./((x-d).^2+h.^2));
% //Combine and return result
Result = 2 * (FirstTerm - SecondTerm)
Result =
Columns 1 through 7:
0.097944 0.133866 0.208270 0.425797 0.692904 -0.140347 -0.124798
Columns 8 and 9:
-0.095581 -0.076166
Related
I am trying to use 'mle' with a custom negative log-likelihood function, but I get the following error:
Requested 1200000x1200000 (10728.8GB) array exceeds maximum array size preference (15.6GB). This might cause MATLAB to become unresponsive.
The data I am using is a 1x1200000 binary array (which I had to convert to double), and the function has 10 arguments: one for the data, 3 known paramenters, and 6 to be optimized. I tried setting 'OptimFun' to both 'fminsearch' and 'fmincon'. Also, optimizing the parameters using 'fminsearch' and 'fminunc' instead of 'mle' works fine.
The problem happens in the 'checkFunErrs' functions, inside the 'mlecustom.m' file (call at line 173, actuall error at line 705).
With 'fminunc' I could calculate the optimal parameters, but it does not give me confidence intervals. Is there a way to circumvent this? Or am I doing something wrong?
Thanks for the help.
T_1 = 50000;
T_2 = 100000;
npast = 10000;
start = [0 0 0 0 0 0];
func = #(x, data, cens, freq)loglike(data, [x(1) x(2) x(3) x(4) x(5) x(6)],...
T_1, T_2, npast);
params = mle(data, 'nloglf', func, 'Start', start, 'OptimFun', 'fmincon');
% Computes the negative log likehood
function out = loglike(data, params, T_1, T_2, npast)
size = length(data);
if npast == 0
past = 0;
else
past = zeros(1, size);
past(npast+1:end) = movmean(data(npast:end-1),[npast-1, 0]); % Average number of events in the previous n years
end
lambda = params(1) + ...
(params(2)*cos(2*pi*(1:size)/T_1)) + ...
(params(3)*sin(2*pi*(1:size)/T_1)) + ...
(params(4)*cos(2*pi*(1:size)/T_2)) + ...
(params(5)*sin(2*pi*(1:size)/T_2)) + ...
params(6)*past;
out = sum(log(1+exp(lambda))-data.*lambda);
end
Your issue is line 228 (as of MATLAB R2017b) of the in-built mle function, which happens just before the custom function is called:
data = data(:);
The input variable data is converted to a column array without warning. This is typically done to ensure that all further calculations are robust to the orientation of the input vector.
However, this is causing you issues, because your custom function assumes data is a row vector, specifically this line:
out = sum(log(1+exp(lambda))-data.*lambda);
Due to implicit expansion, when the row vector lambda and the column vector data interact, you get a huge square matrix per your error message.
Adding these two lines to make it explicit that both are column vectors resolves the issue, avoids implicit expansion, and applies the calculation element-wise as you intended.
lambda = lambda(:);
data = data(:);
So your function becomes
function out = loglike(data, params, T_1, T_2, npast)
N = length(data);
if npast == 0
past = 0;
else
past = zeros(1,N);
past(npast+1:end) = movmean(data(npast:end-1),[npast-1, 0]); % Average number of events in the previous n years
end
lambda = params(1) + ...
(params(2)*cos(2*pi*(1:N)/T_1)) + ...
(params(3)*sin(2*pi*(1:N)/T_1)) + ...
(params(4)*cos(2*pi*(1:N)/T_2)) + ...
(params(5)*sin(2*pi*(1:N)/T_2)) + ...
params(6)*past;
lambda = lambda(:);
data = data(:);
out = sum(log(1+exp(lambda))-data.*lambda);
end
An alternative would be to re-write your function so that it uses column vectors, but you create new row vectors with the (1:N) steps and the concatenation within the movmean. The suggested approach is arguably "lazier", but also robust to row or column inputs.
Note also I've changed your variable name from size to N, since size is an in-built function which you should avoid shadowing.
I am trying to sum very large numbers in MATLAB, such as e^800 and e^1000 and obtain an answer.
I know that in Double-Precision, the largest number I can represent is 1.8 * 10^308, otherwise I get Inf, which I am getting when trying to sum these numbers.
My question is, how do I go about estimating an answer for sums of very, very large numbers like these without using vpa, or some other toolbox?
Should I use strings? It is possible to do this using logs? Can I represent the floats as m x 2^E and if so, how do I take a number such as e^700 and convert it to that? If the number is larger than the threshold for Inf, should I divide it by two, and store it in two different variables?
For example, how would I obtain an approximate answer for:
e^700 + e^800 + e^900 + e^1000 ?
A possible approximation is to use the rounded values of these numbers (I personally used Wolfram|Alpha), then perform "long addition" as they teach in elementary school:
function sumStr = q57847408()
% store rounded values as string:
e700r = "10142320547350045094553295952312676152046795722430733487805362812493517025075236830454816031618297136953899163768858065865979600395888785678282243008887402599998988678389656623693619501668117889366505232839133350791146179734135738674857067797623379884901489612849999201100199130430066930357357609994944589";
e800r = "272637457211256656736477954636726975796659226578982795071066647118106329569950664167039352195586786006860427256761029240367497446044798868927677691427770056726553709171916768600252121000026950958713667265709829230666049302755903290190813628112360876270335261689183230096592218807453604259932239625718007773351636778976141601237086887204646030033802";
e900r = "7328814222307421705188664731793809962200803372470257400807463551580529988383143818044446210332341895120636693403927733397752413275206079839254190792861282973356634441244426690921723184222561912289431824879574706220963893719030715472100992004193705579194389741613195142957118770070062108395593116134031340597082860041712861324644992840377291211724061562384383156190256314590053986874606962229";
e1000r = "197007111401704699388887935224332312531693798532384578995280299138506385078244119347497807656302688993096381798752022693598298173054461289923262783660152825232320535169584566756192271567602788071422466826314006855168508653497941660316045367817938092905299728580132869945856470286534375900456564355589156220422320260518826112288638358372248724725214506150418881937494100871264232248436315760560377439930623959705844189509050047074217568";
% pad to the same length with zeros on the left:
padded = pad([e700r; e800r; e900r; e1000r], 'left', '0');
% convert the padded value to an array of digits:
dig = uint8(char(padded) - '0');
% some helpful computations for later:
colSum = [0 uint8(sum(dig, 1))]; % extra 0 is to prevent overflow of MSB
remainder = mod(colSum, 10);
carry = idivide(colSum, 10, 'floor');
while any(carry) % can also be a 'for' loop with nDigit iterations (at most)
result = remainder + circshift(carry, -1);
remainder = mod(result, 10);
carry = idivide(result, 10, 'floor');
end
% remove leading zero (at most one):
if ~result(1)
result = result(2:end);
end
% convert result back to string:
sumStr = string(char(result + '0'));
This gives the (rounded) result of:
197007111401704699388887935224332312531693805861198801302702004327171116872054081548301452764017301057216669857236647803717912876737392925607579016038517631441936559738211677036898431095605804172455718237264052427496060405708350697523284591075347592055157466708515626775854212347372496361426842057599220506613838622595904885345364347680768544809390466197511254544019946918140384750254735105245290662192955421993462796807599177706158188
Typos fixed from before.
Decimal Approximation:
function [m, new_exponent] = base10_mantissa_exponent(base, exponent)
exact_exp = exponent*log10(abs(base));
new_exponent = floor(exact_exp);
m = power(10, exact_exp - new_exponent);
end
So the value e600 would become 3.7731 * 10260.
And the value 117150 would become 1.6899 * 10310.
To add these two values together, I took the difference between the two exponents and divided the mantissa of the smaller term by it. Then it's just as simple as adding the mantissas together.
mantissaA = 3.7731;
exponentA = 260;
mantissaB = 1.6899;
exponentB = 310;
diff = abs(exponentA - exponentB);
if exponentA < exponentB
mantissaA = mantissaA / (10^diff);
finalExponent = exponentB;
elseif exponentB < exponentA
mantissaB = mantissaB / (10^diff);
finalExponent = exponentA;
end
finalMantissa = mantissaA + mantissaB;
This was important for me as I was performing sums such as:
(Σ ex) / (Σ xex)
From x=1 to x=1000.
I am trying to recreate the following equation in MATLAB
This equation calculates the equivalent inertia.
I have values for Ln, Ln-1 and In which are stored in a matrices and I am assuming that Ltotal is simply the Max value of Ln. Den is a matrix storing the values for the equation.
% Ln
for lp = 1: bars-1
ln_p(1,lp) = radius_pin(1) - radius_pin(lp+1);
ln_w(1,lp) = radius_wheel(1) - radius_wheel(lp+1);
end
% Ln-1
for lp = 1:bars-1
lnMinus_p(1,lp) = radius_pin(1) - radius_pin(lp);
lnMinus_w(1,lp) = radius_wheel(1) - radius_wheel(lp);
end
% L^3 - (Ln-1)^3
lCubed_p = ln_p.^3 - lnMinus_p.^3;
lCubed_w = ln_w.^3 - lnMinus_w.^3;
% In
In_p = aChor_p.^3/12;
In_w = aChor_w.^3/12;
% Denominator of IE equation (inertia)
den_p = lCubed_p./In_p;
den_w = lCubed_w./In_w;
I need a code which will find the Ieq values (they should be in a matrix the same size as the inputs)
If you have an array L which contains all L_n values, and similarly for a vector I with values I_n, then your equation can be implemented simply:
I_eq = L_total^3 / sum( diff(L.^3)./I(2:end) )
I'll also re-write your code, because you seem to be missing the point about MATLAB syntax -- it can be pretty close to mathematical notation. I'll leave out the subscripts (because it's the same calculation for both the pin and wheel anyway):
lCubed = (radius(1)-radius(2:end)).^3 - (radius(1)-radius(1:end-1)).^3;
In = 1/12 * aChor_p.^3;
den = sum(lCubed./In);
I'm struggling with determining the probability of occurrence of unique elements in thresh_strain matrix (which can be seen below as a 100 x 16). I was trying to use the code at the bottom to do this, but I get an equal probability of occurrence associated with each of the elements, whereas I want the probability of occurrence associated with unique elements in thresh_strain.
function [thresh_strain] = MCsolution()
no_iterations = 100;
thresh_strain = zeros(100, 16);
casechoice =input('Enter 1 for 1st Layup and 2 for 2nd layup:');
for i=1:no_iterations
for j=1:16
J = Nielsennew(casechoice);
thresh_strain(i,j) = J(1, j);
end
end
% [uniqueValues,~,uniqueIndex] = unique(thresh_strain);
% frequency = accumarray(uniqueIndex(:),1)./numel(thresh_strain);
Thanks
It is not really clear from the title and description, but I suppose you may be looking for something like this:
myUniqueValues = unique(myMatrix);
nelements = hist(myMatrix(:),myUniqueValues);
%plot(myUniqueValues,nelements)
Basically calculating how often each unique value occurs. From here getting the corresponding percentage is of course trivial.
I am trying to generate an array from some starting values using this formula in MATLAB:
yt = a0 + ∑i=1p (ai ⋅ yt-i), t ≥ p
p is some small number compared to T (max t). I have been able to make this using two for cycles but it is really slow. Is there some easy way to do it?
First p values of y are provided and vector a (its length is p+1) is provided too...
This is what I have so far, but now when I tried it, it doesn't work 100% (I think it's because of indexing from 1 in MATLAB):
y1 = zeros(T+1, 1);
y1(1:p) = y(1:p);
for t = p+1:T+1
value = a1(1);
for j = 2:p+1
value = value + a1(j)*y1(t-j+1);
end
y1(t) = value;
end
EDIT: I solved it, I am just not used to Matlab indexing from 1...
This statement
if(p>=t)
looks odd inside a loop whose index expression is
for t = p+1:T+1
which seems to guarantee that t>p for the entire duration of the loop. Is that what you meant to write ?
EDIT in response to comment
Inside a loop indexed with this statement
for j = 2:p
how does the reference you make to a(j) ever call for a(0) ?
y1 = zeros(T+1, 1);
y1(1:p) = y(1:p);
for t = p+1:T+1
value = a1(1);
for j = 2:p+1
value = value + a1(j)*y1(t-j+1);
end
y1(t) = value;
end