exercise:Ccf27d1eb1: Difference between revisions

From Stochiki
(Created page with "<div class="d-none"> <math> % Generic syms \newcommand\defeq{:=} \newcommand{\Tt}[0]{\boldsymbol{\theta}} \newcommand{\XX}[0]{{\cal X}} \newcommand{\ZZ}[0]{{\cal Z}} \newcomma...")
 
mNo edit summary
 
(One intermediate revision by the same user not shown)
Line 194: Line 194:
Consider the <span class="mw-gls" data-name ="ann">ANN</span> in Figure [[guide:013ef4b5cd#fig_ANN|fig_ANN]] using the <span class="mw-gls" data-name ="relu">ReLU</span> <span class="mw-gls" data-name ="actfun">activation function</span> (see Figure [[guide:013ef4b5cd#fig_activate_neuron|fig_activate_neuron]]).  
Consider the <span class="mw-gls" data-name ="ann">ANN</span> in Figure [[guide:013ef4b5cd#fig_ANN|fig_ANN]] using the <span class="mw-gls" data-name ="relu">ReLU</span> <span class="mw-gls" data-name ="actfun">activation function</span> (see Figure [[guide:013ef4b5cd#fig_activate_neuron|fig_activate_neuron]]).  


Show that there is a particular choice for the weights <math>\weights =(\weight_{1},\ldots,\weight_{9})^{T}</math> such that the resulting hypothesis map <math>h^{(\weights)}(\feature)</math> is a triangle as depicted in Figure [[guide:013ef4b5cd#fig_triangle|fig_triangle]].  
Show that there is a particular choice for the weights <math>\weights =(\weight_{1},\ldots,\weight_{9})^{T}</math> such that the resulting hypothesis map <math>h^{(\weights)}(\feature)</math> is a triangle as depicted in the figure below.


Can you also find a choice for the weights <math>\weights =(\weight_{1},\ldots,\weight_{9})^{T}</math> that produce the same triangle shape if we replace the <span class="mw-gls" data-name ="relu">ReLU</span> <span class="mw-gls" data-name ="actfun">activation function</span> with the linear function <math>\actfun(z) =10 \cdot z</math>?
Can you also find a choice for the weights <math>\weights =(\weight_{1},\ldots,\weight_{9})^{T}</math> that produce the same triangle shape if we replace the <span class="mw-gls" data-name ="relu">ReLU</span> <span class="mw-gls" data-name ="actfun">activation function</span> with the linear function <math>\actfun(z) =10 \cdot z</math>?


<div class="d-flex justify-content-center">
<div class="d-flex justify-content-center">
<span id="fig_triangle"></span>
[[File:fig_triangle.png | 400px | thumb | A hypothesis map <math>h: \mathbb{R} \rightarrow \mathbb{R}</math> with the shape of a triangle. ]]
[[File:fig_triangle | 500px | thumb | A hypothesis map <math>h: \mathbb{R} \rightarrow \mathbb{R}</math> with the shape of a triangle. ]]
</div>
</div>

Latest revision as of 00:03, 13 June 2023

[math] % Generic syms \newcommand\defeq{:=} \newcommand{\Tt}[0]{\boldsymbol{\theta}} \newcommand{\XX}[0]{{\cal X}} \newcommand{\ZZ}[0]{{\cal Z}} \newcommand{\vx}[0]{{\bf x}} \newcommand{\vv}[0]{{\bf v}} \newcommand{\vu}[0]{{\bf u}} \newcommand{\vs}[0]{{\bf s}} \newcommand{\vm}[0]{{\bf m}} \newcommand{\vq}[0]{{\bf q}} \newcommand{\mX}[0]{{\bf X}} \newcommand{\mC}[0]{{\bf C}} \newcommand{\mA}[0]{{\bf A}} \newcommand{\mL}[0]{{\bf L}} \newcommand{\fscore}[0]{F_{1}} \newcommand{\sparsity}{s} \newcommand{\mW}[0]{{\bf W}} \newcommand{\mD}[0]{{\bf D}} \newcommand{\mZ}[0]{{\bf Z}} \newcommand{\vw}[0]{{\bf w}} \newcommand{\D}[0]{{\mathcal{D}}} \newcommand{\mP}{\mathbf{P}} \newcommand{\mQ}{\mathbf{Q}} \newcommand{\E}[0]{{\mathbb{E}}} \newcommand{\vy}[0]{{\bf y}} \newcommand{\va}[0]{{\bf a}} \newcommand{\vn}[0]{{\bf n}} \newcommand{\vb}[0]{{\bf b}} \newcommand{\vr}[0]{{\bf r}} \newcommand{\vz}[0]{{\bf z}} \newcommand{\N}[0]{{\mathcal{N}}} \newcommand{\vc}[0]{{\bf c}} \newcommand{\bm}{\boldsymbol} % Statistics and Probability Theory \newcommand{\errprob}{p_{\rm err}} \newcommand{\prob}[1]{p({#1})} \newcommand{\pdf}[1]{p({#1})} \def \expect {\mathbb{E} } % Machine Learning Symbols \newcommand{\biasterm}{B} \newcommand{\varianceterm}{V} \newcommand{\neighbourhood}[1]{\mathcal{N}^{(#1)}} \newcommand{\nrfolds}{k} \newcommand{\mseesterr}{E_{\rm est}} \newcommand{\bootstrapidx}{b} %\newcommand{\modeldim}{r} \newcommand{\modelidx}{l} \newcommand{\nrbootstraps}{B} \newcommand{\sampleweight}[1]{q^{(#1)}} \newcommand{\nrcategories}{K} \newcommand{\splitratio}[0]{{\rho}} \newcommand{\norm}[1]{\Vert {#1} \Vert} \newcommand{\sqeuclnorm}[1]{\big\Vert {#1} \big\Vert^{2}_{2}} \newcommand{\bmx}[0]{\begin{bmatrix}} \newcommand{\emx}[0]{\end{bmatrix}} \newcommand{\T}[0]{\text{T}} \DeclareMathOperator*{\rank}{rank} %\newcommand\defeq{:=} \newcommand\eigvecS{\hat{\mathbf{u}}} \newcommand\eigvecCov{\mathbf{u}} \newcommand\eigvecCoventry{u} \newcommand{\featuredim}{n} \newcommand{\featurelenraw}{\featuredim'} \newcommand{\featurelen}{\featuredim} \newcommand{\samplingset}{\mathcal{M}} \newcommand{\samplesize}{m} \newcommand{\sampleidx}{i} \newcommand{\nractions}{A} \newcommand{\datapoint}{\vz} \newcommand{\actionidx}{a} \newcommand{\clusteridx}{c} \newcommand{\sizehypospace}{D} \newcommand{\nrcluster}{k} \newcommand{\nrseeds}{s} \newcommand{\featureidx}{j} \newcommand{\clustermean}{{\bm \mu}} \newcommand{\clustercov}{{\bm \Sigma}} \newcommand{\target}{y} \newcommand{\error}{E} \newcommand{\augidx}{b} \newcommand{\task}{\mathcal{T}} \newcommand{\nrtasks}{T} \newcommand{\taskidx}{t} \newcommand\truelabel{y} \newcommand{\polydegree}{r} \newcommand\labelvec{\vy} \newcommand\featurevec{\vx} \newcommand\feature{x} \newcommand\predictedlabel{\hat{y}} \newcommand\dataset{\mathcal{D}} \newcommand\trainset{\dataset^{(\rm train)}} \newcommand\valset{\dataset^{(\rm val)}} \newcommand\realcoorspace[1]{\mathbb{R}^{\text{#1}}} \newcommand\effdim[1]{d_{\rm eff} \left( #1 \right)} \newcommand{\inspace}{\mathcal{X}} \newcommand{\sigmoid}{\sigma} \newcommand{\outspace}{\mathcal{Y}} \newcommand{\hypospace}{\mathcal{H}} \newcommand{\emperror}{\widehat{L}} \newcommand\risk[1]{\expect \big \{ \loss{(\featurevec,\truelabel)}{#1} \big\}} \newcommand{\featurespace}{\mathcal{X}} \newcommand{\labelspace}{\mathcal{Y}} \newcommand{\rawfeaturevec}{\mathbf{z}} \newcommand{\rawfeature}{z} \newcommand{\condent}{H} \newcommand{\explanation}{e} \newcommand{\explainset}{\mathcal{E}} \newcommand{\user}{u} \newcommand{\actfun}{\sigma} \newcommand{\noisygrad}{g} \newcommand{\reconstrmap}{r} \newcommand{\predictor}{h} \newcommand{\eigval}[1]{\lambda_{#1}} \newcommand{\regparam}{\lambda} \newcommand{\lrate}{\alpha} \newcommand{\edges}{\mathcal{E}} \newcommand{\generror}{E} \DeclareMathOperator{\supp}{supp} %\newcommand{\loss}[3]{L({#1},{#2},{#3})} \newcommand{\loss}[2]{L\big({#1},{#2}\big)} \newcommand{\clusterspread}[2]{L^{2}_{\clusteridx}\big({#1},{#2}\big)} \newcommand{\determinant}[1]{{\rm det}({#1})} \DeclareMathOperator*{\argmax}{argmax} \DeclareMathOperator*{\argmin}{argmin} \newcommand{\itercntr}{r} \newcommand{\state}{s} \newcommand{\statespace}{\mathcal{S}} \newcommand{\timeidx}{t} \newcommand{\optpolicy}{\pi_{*}} \newcommand{\appoptpolicy}{\hat{\pi}} \newcommand{\dummyidx}{j} \newcommand{\gridsizex}{K} \newcommand{\gridsizey}{L} \newcommand{\localdataset}{\mathcal{X}} \newcommand{\reward}{r} \newcommand{\cumreward}{G} \newcommand{\return}{\cumreward} \newcommand{\action}{a} \newcommand\actionset{\mathcal{A}} \newcommand{\obstacles}{\mathcal{B}} \newcommand{\valuefunc}[1]{v_{#1}} \newcommand{\gridcell}[2]{\langle #1, #2 \rangle} \newcommand{\pair}[2]{\langle #1, #2 \rangle} \newcommand{\mdp}[5]{\langle #1, #2, #3, #4, #5 \rangle} \newcommand{\actionvalue}[1]{q_{#1}} \newcommand{\transition}{\mathcal{T}} \newcommand{\policy}{\pi} \newcommand{\charger}{c} \newcommand{\itervar}{k} \newcommand{\discount}{\gamma} \newcommand{\rumba}{Rumba} \newcommand{\actionnorth}{\rm N} \newcommand{\actionsouth}{\rm S} \newcommand{\actioneast}{\rm E} \newcommand{\actionwest}{\rm W} \newcommand{\chargingstations}{\mathcal{C}} \newcommand{\basisfunc}{\phi} \newcommand{\augparam}{B} \newcommand{\valerror}{E_{v}} \newcommand{\trainerror}{E_{t}} \newcommand{\foldidx}{b} \newcommand{\testset}{\dataset^{(\rm test)} } \newcommand{\testerror}{E^{(\rm test)}} \newcommand{\nrmodels}{M} \newcommand{\benchmarkerror}{E^{(\rm ref)}} \newcommand{\lossfun}{L} \newcommand{\datacluster}[1]{\mathcal{C}^{(#1)}} \newcommand{\cluster}{\mathcal{C}} \newcommand{\bayeshypothesis}{h^{*}} \newcommand{\featuremtx}{\mX} \newcommand{\weight}{w} \newcommand{\weights}{\vw} \newcommand{\regularizer}{\mathcal{R}} \newcommand{\decreg}[1]{\mathcal{R}_{#1}} \newcommand{\naturalnumbers}{\mathbb{N}} \newcommand{\featuremapvec}{{\bf \Phi}} \newcommand{\featuremap}{\phi} \newcommand{\batchsize}{B} \newcommand{\batch}{\mathcal{B}} \newcommand{\foldsize}{B} \newcommand{\nriter}{R} [/math]

Consider the ANN in Figure fig_ANN using the ReLU activation function (see Figure fig_activate_neuron).

Show that there is a particular choice for the weights [math]\weights =(\weight_{1},\ldots,\weight_{9})^{T}[/math] such that the resulting hypothesis map [math]h^{(\weights)}(\feature)[/math] is a triangle as depicted in the figure below.

Can you also find a choice for the weights [math]\weights =(\weight_{1},\ldots,\weight_{9})^{T}[/math] that produce the same triangle shape if we replace the ReLU activation function with the linear function [math]\actfun(z) =10 \cdot z[/math]?

A hypothesis map [math]h: \mathbb{R} \rightarrow \mathbb{R}[/math] with the shape of a triangle.