【complex】 No.32 support complex for softsign #57552
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你的PR提交成功，感谢你对开源项目的贡献! |
GGBond8488
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Sep 21, 2023
| @@ -2078,6 +2078,30 @@ struct SoftsignGradFunctor : public BaseActivationFunctor<T> { | |||
| static constexpr ActBwdOpFwdDeps FwdDeps() { return ActBwdOpFwdDeps::kDepX; } | |||
| }; | |||
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| // x = a+bi | |||
| // |x| = r = sqrt(a^2 + b^2) | |||
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可以试试两个思路，1. |x| = sqrt(x * x.conj)，用这个思路往下推
2. 通过链式法则的方式反推回来 out = x.divide(1+abs(x)), 求dout和dx的关系
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请问一下第一点是关于|x|的微分公式吗，还是就是x的复数绝对值。
第二点用链式法则反推，我发现最后的结果还是跟我直接微分的表达式一样。。。请问有什么方法可以验证一下求导的公式对错吗
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- 请问一下第一点是关于|x|的微分公式吗，还是就是x的复数绝对值。
复数的abs你可以理解为求模，刚好可以等价为sqrt(x*x.conj)，可以把这个带入进去求导 - 链式法则反推，我发现最后的结果还是跟我直接微分的表达式一样 。
｜x｜在复数场景下比较特殊，属于文档中C->R的函数，比较特殊，所以你会发现直接用abs的反向会计算不对，至于怎么验证是否正确，一般能过单测就应该是算对了
There was a problem hiding this comment.
- 请问一下第一点是关于|x|的微分公式吗，还是就是x的复数绝对值。
复数的abs你可以理解为求模，刚好可以等价为sqrt(x*x.conj)，可以把这个带入进去求导- 链式法则反推，我发现最后的结果还是跟我直接微分的表达式一样 。
｜x｜在复数场景下比较特殊，属于文档中C->R的函数，比较特殊，所以你会发现直接用abs的反向会计算不对，至于怎么验证是否正确，一般能过单测就应该是算对了
请问一下abs求导的时候要根据sqrt(x*x.conj)这个式子求导吗，一直算出来不对
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@GGBond8488 @luotao1 麻烦review一下，谢谢~ |
|
重新提个commit触发一下coverage ci 吧 |
#58545 |
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Description
support complex for softsign
#56145