Christopher A. Wong
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 Apr 19 revised Local Truncation Error of Implicit Euler added expanded answer Apr 19 answered Local Truncation Error of Implicit Euler Apr 19 comment Use trigonometric polynomial to approximate periodic function. OK, I see understand your concern now; it is my mistake. I misread your question. It's not the partial sum that's converging uniformly, but the Cesaro mean. It is false that the partial sums converge uniformly if $f$ is continuous. Apr 18 comment Use trigonometric polynomial to approximate periodic function. I don't think you're missing anything. Apr 6 comment Prove that if $x,y$ are vectors in some unitary space and satisfy $||x||=||y||$ then $\exists T\in Hom(V,V),T^*=T^{-1}\land T(x)=y$ To elaborate on the above hint, you might imagine a matrix representation of $T$ as being block-diagonal with a $2 \times 2$ block that captures the $x \mapsto y$ behavior, and then the rest of the matrix can be whatever you like to keep it unitary. Apr 5 revised Is a continuous function plus a discontinuous function discontinuous? latex Apr 4 comment Correct way to solve a limit. Your expression in Step 1 is not correct. Also, limits are never applied by setting $n = \infty$. Surely your textbook does not do it this way, right? Mar 29 answered Dimension of Basis of Subspace Mar 23 awarded Nice Answer Mar 22 answered Are there any objects which aren't sets? Mar 2 awarded Enlightened Mar 2 awarded Nice Answer Mar 1 comment Does $X((X^TX)^{-1}X^Ty)$ equal $X(X^{-1}y)$? @user318895 Why are you computing $XX^{-1} y$? If $X$ is indeed square and invertible, then $XX^{-1} = I$. If you're actually doing $XX^{+}y$, then yes, Matlab knows more efficient ways to compute $X^{+}$ than to do $(X^T X)^{-1} X^T$. Mar 1 revised Does $X((X^TX)^{-1}X^Ty)$ equal $X(X^{-1}y)$? added 183 characters in body Mar 1 comment Does $X((X^TX)^{-1}X^Ty)$ equal $X(X^{-1}y)$? @Hugo It's not. Your answer is correct. But in the context that the OP mentioned, what should be clarified is that the least-squares projector $(X^T X)^{-1} X^T$ gives the general solution, while $X^{-1}$ is equal to this only when $X$ is indeed square. Mar 1 comment Does $X((X^TX)^{-1}X^Ty)$ equal $X(X^{-1}y)$? @Jonas I don't think so. At face value, e.g. the title of the question, yes, but really the OP's question is "Why does someone use $X(X^{-1}y)$ instead of $X(X^T X)^{-1} X^{T}y$"? And the answer to that question is that one is a special case of the other. Mar 1 comment Does $X((X^TX)^{-1}X^Ty)$ equal $X(X^{-1}y)$? @Jonas, I think in the OP's case, it must be an abuse of notation; if $X^{-1}$ existed then there would be no need for considering a least-squares solution. Otherwise what would be the point of even writing $XX^{-1}$ when you could just write $I$. Mar 1 answered Does $X((X^TX)^{-1}X^Ty)$ equal $X(X^{-1}y)$? Mar 1 comment Prove that $R(T^{*})^\perp =N(T)$ @George, I have modified the symbols to indicate how it works for the general case $T: V \rightarrow W$. In particular, the proof is not different, you just have to know which space, and what inner product (whether on $W$ or $V$), you're working with. Mar 1 revised Prove that $R(T^{*})^\perp =N(T)$ modified solution for the general case $V \rightarrow W$.