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Okay this may sound stupid but i need a little help.what do $\\large \\frac{d}{dx}$ and $\\large \\frac{dy}{dx}$ mean $$\int {x^xdx} = \int {e^ {\ln x^x}dx} = \int {\sum_ {k=0}^ {\infty}\frac {x^k\ln. I need a thorough explanation
Rankeya has given a valid answer to the written question, but i realize now i was too vague So there is no simple answer to your question, unless you are willing to consider a series approximation, obtained by expanding the exponential as a series Secondly, i looked up the correct exercise in jacobson and found that the following exercise is precisely to show that it does hold for all division rings
Stupid gut feelings.i'm accepting this answer and reposting the correct question.
This isn't really an answer as it stands Answer should be self contained, but this answer lacks the mathematically acceptable explanation it alludes to, so it's not very useful If you're going to reproduce anything from a book, you might as well reproduce the part that answers the question rather than the part that asks the questions. Expanding out $ (x-y)^2$, we get $$ x^2+ (x-y)^2+y^2=2 (x^2-xy+y^2)$$ Then complete the square: $$ x^2-xy+y^2=\Big (x-\frac {y} {2}\Big)^2+\frac {3y^2} {4} $$ so that.
Its actually $\dot x$ or $\frac {dx} {dt}$, the term inside of the integral. Eigenvectors corresponding to distinct eigenvalues of a symmetric matrix must be orthogonal to each other Eigenvectors corresponding to the same eigenvalue need not be orthogonal to each other However, since every subspace has an orthonormal basis, you can find orthonormal bases for each eigenspace, so you can find an orthonormal basis of eigenvectors.
Use a kronecker product to flatten the matrix to a vector, i.e
$\,w\to w.\,$ then the equation becomes $ (x\otimes i)^tw +b$ whose gradient is $ (x\otimes i)$. Explain $\\iint \\mathrm dx\\mathrm dy = \\iint r \\mathrm d\\alpha\\mathrm dr$ i'm reading the proof of gaussian integration When we change to polar coordinates, why do we get an As noted in the comments, your derivation contains a mistake
To answer the question, this function can not be integrated in terms of elementary functions
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