Recent content by Passers_by

OK

Would you consider selling Moise's calculus

Thank you. In my current situation, I am fit to buy e-books. It's a pity these books are published so early.

https://www.amazon.com/dp/B000L3UO2A/?tag=pfamazon01-20 I don't know where to buy it, and I hope I can buy it as an ebook.

What about the Calculus by Edwin E. Moise? Is it better than Thomas's 3rd ed Calculus? I found two editions of Moise's calculus, one in 1967 and the other in 1972.

I don't know how do I get the average separation in the spherical model.

The reason for the confusion is that my original idea was to prove that the subspace normal to ##n_3## is a eigenspace with eigenvalue equal to ##\lambda_1##. I'm not taking this as a given premise.

I might know what you mean. If ##v_2=k_1n_1+k_2n_2##, then ##Sv_2=\lambda_1v_2##. I knew that from the beginning. And it all comes with a premise: there are two eigenvectors(##n_1##, ##n_2##) that are orthogonal to each other in the subspace in which any vector is normal to ##n_3## and their...

##n_3\cdot v_2=k_3##. But there's something wrong with that proof in #15. Please don't be bothered by that.

Here is my idea of proof. First, prove that ##v_2## is an eigenvector. You can see that in #13. And then let's prove that its eigenvalue is ##\lambda_1##. And to prove that, I found that any eigenvector with an eigenvalue of ##\lambda_3## is linearly dependent. So ##\lambda_3## is not the...

I rewrote the proof as follows: Expand ##n_3## with eigenvalue equal to ##\lambda_3## in an orthogonal base formed by eigenvectors: ##n_1^*,n_2^*,n_3^*##(##n_1^*,n_2^* ##with eigenvalue equal to ##\lambda_1,n_3^*## with eigenvalue equal to ##\lambda_3##)$$n_3=k_1n_1^*+k_2n_2^*+k_3n_3^*,$$...

Expand ##v_2## in an orthogonal base formed by eigenvectors: ##n_1,n_2,n_3,##$$v_2=k_1n_1+k_2n_2+k_3n_3,$$$$Sv_2=\lambda_1\left(k_1n_1+k_2n_2\right)+\lambda_3k_3n_3,$$but the subspace spanned by ##n_1,n_2## and the subspace by ##n_3## do not intersect, so either ##k_1=k_2=0## or ##k_3=0##. if...

##n_3## is a eigenvector with eigenvalue equal to ##\lambda_3##.Because ##\lambda_1 \ne \lambda_3##, so there is a eigenvector ##n_1## with eigenvalue equal to ##\lambda_1## and ## n_1 \perp n_3##. Let ##v_2 \perp n_1## and ##v_2 \perp n_3##,next we can prove ##v_2## is a eigenvalue too. For a...

Any vector which is normal to n3 is then an eigenvector of S with eigenvalue equal to λ1.

I don't think there's much difference between what you're saying and the method I use:confused: