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Old 08-17-2012, 05:56 PM
jianpan jianpan is offline
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Default backpropagation at the final layer

since \delta_1^{(L)}=\frac{\partial e(w)}{\partial s_1^{(L)}}, and e(w)=(x_1^{(L)}-y_n)^2, and x_1^{(L)}=\theta(s_1^{(L)}), I got \delta_1^{(L)}=2*(x_1^{(L)}-y_n)*(1-\theta^2(s)). is this correct? I found a python program at wikipedia about backpropagation using the \delta_1^{(L)}=(x_1^{(L)}-y_n)*(1-\theta^2(s)), and couldn't figure out where the 2 was dropped. Can someone confirm my derivation is correct? Thanks.
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Old 08-17-2012, 08:12 PM
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yaser yaser is offline
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Default Re: backpropagation at the final layer

Quote:
Originally Posted by jianpan View Post
since \delta_1^{(L)}=\frac{\partial e(w)}{\partial s_1^{(L)}}, and e(w)=(x_1^{(L)}-y_n)^2, and x_1^{(L)}=\theta(s_1^{(L)}), I got \delta_1^{(L)}=2*(x_1^{(L)}-y_n)*(1-\theta^2(s)). is this correct? I found a python program at wikipedia about backpropagation using the \delta_1^{(L)}=(x_1^{(L)}-y_n)*(1-\theta^2(s)), and couldn't figure out where the 2 was dropped. Can someone confirm my derivation is correct? Thanks.
Some people define the error with a factor of 1 \over 2 in it in anticipation of the differentiation, hence the discrepancy.
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Old 08-18-2012, 12:05 AM
jianpan jianpan is offline
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Default Re: backpropagation at the final layer

Thanks for the clarification, Professor.
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