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I need to prove the idendity of $$\delta(g(x)) = \Sigma \frac{\delta(x-x_i)} {|g'(x_i)|}$$ and we know that $$g(x_i) = 0$$

So I think we can write,

$$g(x) = (x - x_1)(x-x_2)...(x-x_i)$$

so we are actually trying to find,

$$\delta((x - x_1)(x-x_2)...(x-x_i))$$

Now I am not sure what to do. I tried to write $g(x)$ in terms of $g'(x)$ to catch some sort of similarity but it did not work out well.

In another approach, I tried to make a taylor expansion around $x_i$ which gives something like

$$g(x = x_i) = g(x_i) + (x - x_i)g'(x_i) + ...$$

(In here I ignored the higer terms with no reason actually)

Hence around $x_i$, $\delta(g(x)) = \frac{\delta(x-x_i)} {|g'(x_i)|}$ (by using $\delta(g'(x_i) (x-x_i)) = \frac{\delta(x-x_i)} {|g'(x_i)|}$)

since we have $i$ points,

$$\delta(g(x)) = \Sigma \frac{\delta(x-x_i)} {|g'(x_i)|}$$

Is this looks stupid ?

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    $\begingroup$ Yes. That $g$ is smooth with only finitely many zeros $x_i$ where $g'(x_i)\ne 0$ means $g(x) = (x - x_1)(x-x_2)...(x-x_i) h(x)$ with $h$ smooth non-vanishing and $h'(x_i) = \frac{g'(x_i)}{\prod_{j \ne i} (x_i-x_j)}$. Then you didn't define $\delta(g(x))$ so there is quite nothing to prove. $\endgroup$ – reuns Oct 9 at 18:02
  • $\begingroup$ @reuns I did not understand much. Hmm are you meaning that I forget h(x) so I am doing things wrong ? $\endgroup$ – Reign Oct 9 at 18:24
  • $\begingroup$ Sure, once we know the sign of $g(x_1-1)$ only the sign of the $g'(x_i)$ are constrained but $|g'(x_i)|$ can be anything. $\endgroup$ – reuns Oct 9 at 18:30
  • $\begingroup$ I still cant do it $\endgroup$ – Reign Oct 9 at 18:52
  • $\begingroup$ I meant $h(x_i) = \frac{g'(x_i)}{\prod_{j \ne i} (x_i-x_j)}$ above. Do what ? $\endgroup$ – reuns Oct 9 at 19:25

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