assignment3

# Plugging these in one 12y 1 t can simplify the

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Unformatted text preview: hown ˆ that the minimum length solution to y =� A, x �, is x = A � A, A �−1 y . So, for our problem, �� � �� �� � −1 y x= T −t 1 � T −t 1 , T −t 1 � ˆ . 0 Where, using the deﬁnition of the Grammian, we have that: � � � �� � &lt; T − t, T − t &gt; &lt; T − t, 1 &gt; � T − t 1 , T − t 1 �= . &lt; 1, T − t &gt; &lt; 1, 1 &gt; 2 Now, we can use the deﬁnition for inner product to ﬁnd the individual entries, &lt; T − t, T − t &gt;= � T �T 2 3 2 0 (T − t) dt = T /3, &lt; T − t, 1 &gt;= 0 (T − t)dt = T /2, and &lt; 1, 1 &gt;= T . Plugging these in, one 12y 1 t can simplify the expression for x and obtain x(t) = T 2 [ 2 − T ] for t ∈ [0, T ]. ˆ ˆ Alternatively, we have that x(t) = p�t). Integrating b� ¨ oth sides and taking into account that � t ( t1 t ˙(0) = 0, we have p(t) = 0 0 x(τ )dτ dt1 = 0 f (t1 )dt1 . Now, we use the integration p(0) = 0 and p �t �t �t by parts formula, 0 u dv = uv |t − 0 v du, with u = f (t1 ) = 0 1 x(τ )dτ, and dv = dt1 ; hence du = 0 �t�t df (t1 ) = x(t1 )dt1 and v = t1 . Plugging in and simplifying we get that p(t) = 0 0 1 x(τ )dτ dt1 = � t �T ) 0 (t − τ )x(τ� dτ. Thus, y = p(T ) = 0 (T − τ )x(τ )dτ =&lt; T − t, x(t) &gt; . In addition, we have that T 0 = p T ) = 0 x(τ )dτ =&lt; 1, x(t) &gt; . That is, we seek to ﬁnd the minimum length x(t) such that ˙( y = &lt; T − t, x(t) &gt; 0 = &lt; 1, x(t) &gt; . Recall that the minimum length solution x(t) must be a linear combination of T − t and 1, i.e., ˆ x(t) = a1 (T − t) + a2 . So, ˆ y = &lt; T − t, a1 (T − t) + a2 &gt; = a1 0= &lt; 1, a1 (T − t) + a2 &gt; = �T 0 �T 3 2 (T − t)2 dt + a2 0 (T − t)dt = a1 T3 + a2 T2 �T 2 = a1 T2 + a2 T. 0 (a1 (T − t) + a2 )...
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