Mathematical Analysis - Equivalents Part 1 and 2 Solutions

Extract

[...] Thus, the function is differentiable on . The only delicate point is in . We are interested in the partial derivative with respect to , on a so . For what concerns that in or on a so we deduce soit then so finally . So . So we have . All class functions est differentiable on . Exercise The function is differentiable on and the partial derivatives are respectively: We are looking for when these partial derivatives cancel out or . [...]

[...] Mathematical Analysis - Equivalents Part Exercise Let : Sur on a so and more so we deduce by multiplication of positive terms that and for comparison with the Riemann series the series converges. On a . Thus, as , we deduce the following result therefore, according to the rule of Alembert, the series converges. We have a general term of the form with so according to Bertrand's series the series is divergent. The sequence is monotone (decreasing) and hence according to the special criterion of alternating series, the series converges. [...]

[...] We then find that . Or we have a , and . Thus, these critical points are not extrema and the function does not admit it. Exercise We have the following partial derivatives: I don't know what they are and . On So On We are looking soit . We then obtain by injection so the same what ultimately gives then and . We also obtain then by injection so then what ultimately gives then et . On a alors . [...]

[...] Finally, we deduce that for all, on a sum up these relations for à (by linearity of the integral) : and like and that (car on this interval) it comes what we must prove is: On the one hand , on the other hand . In addition, we have the middle term which is worth . On a and so like and . Thus, by encadrement we have be it by definition . Part Exercise The function is continuous everywhere except at the origin immediately. By passing to polar coordinates one can take and . Thus, we find . [...]

[...] Finally using the linearity of the sum . In addition, we can calculate the value of the integral , it follows that so . We want such as searching for the integer such as , we find . As the previous function here associate here is decreasing and so exactly the same development that leads to . Passing to the limit . We recognize a form the integral thus we have . We are looking , we are looking then . [...]