How to multiply symbolic with numeric?

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Ali Najem 2020-12-21

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https://ww2.mathworks.cn/matlabcentral/answers/698975-how-to-multiply-symbolic-with-numeric

评论： Ali Najem 2020-12-23

采纳的回答： Steven Lord

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Hi all'

assume that

w=randn(2,2)
a=sym('a',[2 2])

I need to get the answer of multiblication of (a .* w) with smae size [ 2 2] since, I multiplied them like (a .* w) i got this answer

ans =

[ -(4545034027795199*a1_1)/18014398509481984, -(8311781718565711*a1_2)/18014398509481984]

[ -(1390961227654289*a2_1)/1125899906842624, (5388115603050723*a2_2)/18014398509481984]

However, the random values of w is

w =

-0.2523 -0.4614

-1.2354 0.2991

a =

[ a1_1, a1_2]

[ a2_1, a2_2]

I don't belive this correct results however, I need the results like this [ -0.2523 * a1_1, -0.4614* a1_2

-1.2354 * a2_1, 0.2991 * a2_2]

thanks indeed

采纳的回答

Steven Lord 2020-12-21

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Call vpa on the result.

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Walter Roberson 2020-12-21

vpa(w.*a,3)

Ali Najem 2020-12-21

Thnks indeed sir

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Walter Roberson 2020-12-21

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在 MATLAB Online 中打开

w = round(sym(randn(2,2),'d'),4)

You are, by the way, not correct about what the value of w is. The actual values for w extend to about 15 significant decimal places, and you are getting confused because you have the default format in effect. I recommend that you use

format long g

and change your preferences to use that.

However, you defined your required output in terms of four decimal places, so I included a call to force four decimal places.

Note: symbolic floating point numbers give the strong impression that they are base 10 internally, but closer to the truth is that they use a base 2^30 representation, which is close enough to base 10^10 representation that it takes real effort to prove the difference.

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Ali Najem 2020-12-22

编辑：Ali Najem 2020-12-22

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Alright, sir I need to generate code of neural network feedforwad and also backword peocesses in symbolic form. see the example

w12=sym('w12',[80,784]);
w23=sym('w23',[60,80]);
w34=sym('w34',[10,60]);
b12=sym('b12',[80,1]);
b23=sym('b23',[60,1]);
b34=sym('b34',[10,1]);
cox1=sym('cox1',[80,784]);
cox2=sym('cox2',[60,80]);
cox3=sym('cox3',[10,60]);
cox4=sym('cox4',[80,1]);
cox5=sym('cox5',[60,1]);
cox6=sym('cox6',[10,1]);
a1=sym('a1',[784,1]);
y=sym('y',[10,1]);
z2=w12*a1+b12;
a2 = z2./(1+exp(-z2));  
z3=w23*a2+b23;
a3= z3./(1+exp(-z3));
z4=w34*a3+b34;
a4 = z4./(1+exp(-z4));
MSE= 1/2.*(a4-y).^2;
dMSEda4=(a4-y);
dL4= a4+ 1./(1+exp(-z4)).*(1- a4);
dL3= a3+ 1./(1+exp(-z3)).*(1- a3);
dL2= a2+ 1./(1+exp(-a2)).*(1- a2);
error4 = dMSEda4.* dL4;
error3 = (w34.'*error4).*dL3;
error2 = (w23.'*error3).*dL2;    
errortot4 =  error4;
errortot3 =  error3;
errortot2 =  error2;
grad4 =  error4*a3.';
grad3 =  error3*a2.';
grad2 =  error2*a1.';
%%
mse=sum(MSE);
cox1_grad2= sum(sum(cox1.*grad2));
cox2_grad3= sum(sum(cox2.*grad3));
cox3_grad4= sum(sum(cox3.*grad4));
cox4_errortot2= sum(sum(cox4.*errortot2));
cox5_errortot3= sum(sum(cox5.*errortot3));
cox6_errortot4= sum(sum(cox6.*errortot4));
H= mse+ cox1_grad2+ cox2_grad3+ cox3_grad4+ cox4_errortot2+ cox5_errortot3+ cox6_errortot4;
%% dervative of W12  
for i=1:80
    for j=1:784
        dH_dw12(i,j)=diff(H,w12(i,j));
    end
end 
%% dervative of W23  
for i=1:60
    for j=1:80
        dH_dw23(i,j)=diff(H,w23(i,j));
    end
end 
%% dervative of W34  
for i=1:10
    for j=1:60
        dH_dw34(i,j)=diff(H,w34(i,j));
    end
end 
%% dervative of b12
for i=1:80
    for j=1:1        
     dH_db12(i,j)=diff(H,b12(i,j));
    end
end 
%% dervative of b23
for i=1:60
    for j=1:1        
     dH_db23(i,j)=diff(H,b23(i,j));
    end
end 
%% dervative of b34
for i=1:10
    for j=1:1
        dH_db34(i,j)=diff(H,b34(i,j));
    end
end 

This is it what am try to do obvesily, when i run this code I didn't get any results at all even after 5 hours of running.

Really appreciate your help.

Walter Roberson 2020-12-23

Taking one derivative of one entry of dL4 took my system about 18 minutes. You have over 65000 derivatives to take at the bottom of your code, each on a value that is more complicated than dL4 is. You would expect a minimum of 20 minutes per derivative, times 65000 derivivates, gives about 130000 minutes -> over 900 days.

Each derivative looks like it is going to be over 2 gigabytes to write out -> 130 petabytes .

Suppose that I whipped up something to run on distributed AWS and (somehow!!) managed to create that 130+ petabyte file tomorrow: what would you do with it?

Ali Najem 2020-12-23

It's just an idea for a project, anyway your explanation is clear and I have understood all things thank you so much sir really appreciate your time to help.

Maybe I will try another idea,

my regards.

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