@@ -140,6 +140,7 @@ pour :math:`\lambda` et *z* en faisant tendre *h* vers 0.
140140On commence par le plus simple, le cas float 8 pour lequel on impose :math: `z=0 `.
141141
142142.. math ::
143+ :label: eq-qua-1
143144
144145 f(B,\lambda ,h) = \frac {1 }{h} \sum _{k=1 }^{n} \sum _{ij} \pa {b_{ij} - \lambda d_k - z }^2
145146 K\pa {\frac {b_{ij} - \lambda d_k - z}{h}}
@@ -154,3 +155,43 @@ ce calcul devient une somme d'espérence.
154155
155156 Résolution
156157==========
158+
159+ If :math: `K(u)=\frac {1 }{\sqrt {2 \pi }}e^{-\frac {1 }{2 }u^2 }` then
160+ :math: `K'(u) = -u \frac {1 }{\sqrt {2 \pi }}e^{-\frac {1 }{2 }u^2 } = -u K(x)`.
161+ Let's denote :math: `g(b,x) = (b-xd)^2 K\pa {\frac {b-xd}{h}}`. Then:
162+
163+ .. math ::
164+
165+ \begin {array}{rcl}
166+ g(b,x) &=& \frac {1 }{h} (b-xd)^2 K\pa {\frac {b-xd}{h}} \\
167+ \frac {\partial g}{\partial x}(b,x) &=&
168+ \frac {1 }{h}\cro { -2 d(b-xd)K\pa {\frac {b-xd}{h}} -\frac {d}{h} (b-xd)^2 K'\pa {\frac {b-xd}{h}} } \\
169+ &=& -\frac {d(b-xd)}{h}\cro {2 K\pa {\frac {b-xd}{h}} + \frac {b-xd}{h} K'\pa {\frac {b-xd}{h}} }
170+ \end {array}
171+
172+ Applied to :eq: `eq-qua-1 `:
173+
174+ .. math ::
175+
176+ \begin {array}{rcl}
177+ f(B,\lambda ,h) &=& \frac {1 }{h} \sum _{k=1 }^{n} \sum _{ij} \pa {b_{ij} - \lambda d_k}^2
178+ K\pa {\frac {b_{ij} - \lambda d_k}{h}} \\
179+ &=& \sum _{k=1 }^{n} \sum _{ij} g(b_{ij}, \lambda )
180+ \end {array}
181+
182+ Then:
183+
184+ .. math ::
185+
186+ \begin {array}{rcl}
187+ \frac {\partial f}{\partial \lambda } &=& \sum _{k=1 }^{n} \sum _{ij}
188+ \frac {\partial g}{\partial \lambda }(b_{ij}, \lambda )
189+ \end {array}
190+
191+ Notebooks
192+ =========
193+
194+ .. toctree ::
195+ :maxdepth: 1
196+
197+ ../notebooks/dsgarden/quantization_f8
0 commit comments