diff --git a/ea/ubA/EAA10HohlochMutschler.pdf b/ea/ubA/EAA10HohlochMutschler.pdf
new file mode 100644
index 0000000..1ee9b4d
--- /dev/null
+++ b/ea/ubA/EAA10HohlochMutschler.pdf
Binary files differ
diff --git a/is/ub9/is9.pdf b/is/ub9/is9.pdf
index cd12f41..d944551 100644
--- a/is/ub9/is9.pdf
+++ b/is/ub9/is9.pdf
Binary files differ
diff --git a/is/ub9/is9.tex b/is/ub9/is9.tex
index e1c11b7..c34f421 100644
--- a/is/ub9/is9.tex
+++ b/is/ub9/is9.tex
@@ -85,11 +85,25 @@
         Da $Y$ qudratisch von $X$ abhängt, vergrößert sich auch die Varianz. Entsprechend werden die Residuen für größere $Y$ ebenfalls größer. %TODO: ?
     \Aufgabe{Kernels}{50}
         \begin{enumerate}
-            \item $k(x_1,x_2)=C=\sqrt{C}^2\cdot k'(x_1,x_2)$, where $k'(x_1,x_2)=1$ which is clearly positive semidefinite.
-            \item is p.d. according to the lecture, because it's a polynomial kernel
-            \item TODO: don't see why not in the hint-case
-            \item As 2. is a kernel, this is too with $\alpha=\sqrt{5}$
-            \item As $k_1(x_1,x_2)=x_1^Tx_2$ and $k_2(x_1,x_2)=1$ are kernels, their sum is too. Now we multiply the sum of this kernels with itself and get another kernel.
+            \item $k(x_1,x_2)=C= \alpha \cdot k'(x,y)\\
+            k'(x,y)=1 \begin{pmatrix}
+            x &y
+            \end{pmatrix} \begin{pmatrix}
+            1 &1\\1&1
+            \end{pmatrix}\begin{pmatrix}
+            x\\y 
+            \end{pmatrix}= x^2+2xy+y^2=(x+y)^2 \geq 0 \forall{x,y}\\$
+            \item $k(x,y) =xy\\
+             <x,y> \geq 0 \forall{x,y} $
+            \item No kernel $k(x,y) =\dfrac{1}{2}(x+y)\\
+            k(-1,-1) =-1 \ngeq 0\\$
+            But for a Kernel ist has to be:\\
+            $k(x,x) \geq 0 \forall{x} $     
+            \item $k(x,y) = 5x^Ty = 5 \cdot k'(x,y)$ with $k'(x,y) = x^Ty =<x,y>$\\
+            $= \alpha *k'(x,y)$ where $\alpha =5$  
+            \item $k(x,y)=(x^Ty+1)^2= (k_1(x,y)+k_2(x,y))^2$\\
+            where $k_1(x,y)=x^Ty, k_2(x,y)=1$\\
+            $k_1$ and $k_2$ are proven as kernels in task 2 and 1.
         \end{enumerate}
 \end{document}
 
diff --git a/mr/ub9/loc_framework/localize_pf.m b/mr/ub9/loc_framework/localize_pf.m
index f9ce2c6..7160621 100644
--- a/mr/ub9/loc_framework/localize_pf.m
+++ b/mr/ub9/loc_framework/localize_pf.m
@@ -56,12 +56,14 @@
         pf_w = cat(2,pf_w, currw);
 
         % Resampling step:
-
-        currx = pf_resample_sys(currx,w,ESSmin);
+        ess= 1/sum(arrayfun(@(x) x^2,w));
+        if ess < ESSmin
+            
+            currx = pf_resample_sys(currx,w,ESSmin);
 
         % Do not forget to reset weights after resampling:
-        w = ones(nSamples,1)/nSamples;
-
+            w = ones(nSamples,1)/nSamples;
+        end
     end % for each control input
 
     plot_trajectory_particles(x_true, pf_x, pf_w, dt);
@@ -84,4 +86,7 @@
 %2) ESSmin = 0.4*20;  errpos: 0.6882   erryaw=0.1149
 %3)ESSmin = always > ess;  errpos: 0.5342   erryaw=0.0174
 %4)ESSmin = 0.1*20= errpos= 0.8369  err_yae 0.1608
-%Sollen wir hier die Zeit auch noch messen?
\ No newline at end of file
+%Sollen wir hier die Zeit auch noch messen?
+
+%Gemeint war, dass ich resamplings nur manchmal mache oder nie oder die
+%AArt des resampling.
\ No newline at end of file
diff --git a/mr/ub9/loc_framework/pf_resample_sys.m b/mr/ub9/loc_framework/pf_resample_sys.m
index fc0f86e..95d5831 100644
--- a/mr/ub9/loc_framework/pf_resample_sys.m
+++ b/mr/ub9/loc_framework/pf_resample_sys.m
@@ -3,10 +3,10 @@
 
 % YOUR CODE STARTS HERE:
 
-ess= 1/sum(arrayfun(@(x) x^2,w));
+
 
 M = size(currx, 3);
-if(ess < essMin)
+%if(ess < essMin)% nicht hier sondern in der plocalize pf. Nach dem resambling alle gewichte auf 
 
     nextx = zeros(3, 1, M);
 
@@ -21,9 +21,8 @@
             end
             nextx(:,:,m)=currx(:,:,i);
     end
-else
         nextx=currx;
-end
+
 % YOUR CODE END HERE
 
 end