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I am new to neural networks and recently found out about gradient descent.

Something does not sit right with me.

x←x−λ∇fk(x)

Why does this formula work? Wouldn't it make more sense to have lambda a large value thereby mimizing the cost function?

I am not phrasing my question properly as i am honestly quite confused. How could gradient descent result in a global optimum if it always reduces the value?

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    $\begingroup$ the lambda controls the descent, you can quickly become unstable and finding the solution will be difficult if not impossible. In a similar way that you have to be careful of step size when solving certain differential equations (namely some nonlinear ones) $\endgroup$ – Chinny84 Nov 26 '15 at 15:05
  • $\begingroup$ If you are looking at minizing the function wouldn't it make sense to have infinity as lambda? And x - infinity would give u negative infinity? $\endgroup$ – aceminer Nov 26 '15 at 15:06
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    $\begingroup$ You can adjust $\lambda$ using line search. Have a look at en.wikipedia.org/wiki/Line_search $\endgroup$ – Claude Leibovici Nov 26 '15 at 15:08
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    $\begingroup$ What I am confused about is a case when the loss function actually is not minimized when using a huge learning rate as opposed to a smaller one $\endgroup$ – aceminer Nov 26 '15 at 15:10
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Let me explain you clearly:

Learning rate is the length of the steps the algorithm makes down the gradient on the error curve.

So, in case you have a high learning rate, the algorithm might overshoot the optimal point.

And with a lower learning rate, in case of any overshoot, the magnitude of overshoot would be lesser than when you have a higher learning rate.

So, in case of overshoot, you would end up at a non-optimal point whose error would be higher.

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  • $\begingroup$ Yes and why isit so? Can I have an example of such a case? Maybe some equations and numbers $\endgroup$ – aceminer Nov 27 '15 at 5:53
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    $\begingroup$ You might want to have a watch at this lecture of Professor Andrew Ng. He did a great job in explaining the downside of having high learning rate. $\endgroup$ – Dawny33 Nov 27 '15 at 5:54

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