network profile
| bio | website | kulu.lu |
|---|---|---|
| location | Israel | |
| age | ||
| visits | member for | 8 months |
| seen | 4 hours ago | |
| stats | profile views | 28 |
I am a an entrepreneur and web developer.
Working on my startup - A social video player for groups and friends.
You can check it out on:
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5h |
comment |
Proof of eigenvectors of a rotation matrix in complex plane @Berci Thanks for your response. I cannot see however the trigonometric \ algebraic proof behind that. Thanks1 |
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6h |
comment |
calculate kernels of matrices with angles @wantToLearn do u by any chance study in the Hebrew university ? :) |
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6h |
asked | Proof of eigenvectors of a rotation matrix in complex plane |
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2d |
accepted | If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge? |
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2d |
comment |
Positive series problem fantastic. thanks for referring to the book. |
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2d |
comment |
If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge? @Somabha, Thanks. Trying to translate the concepts to English is very confusing. My apologies |
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2d |
comment |
If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge? Hi John, you are right. My mistake, I meant that the series diverges to +Infinity. I'm very sorry for the inconvenience. Trying to translate the concepts to English is very confusing. My apologies. |
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2d |
revised |
If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge? added 4 characters in body |
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2d |
awarded | Custodian |
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2d |
reviewed | Approve suggested edit on If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge? |
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2d |
asked | If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge? |
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May 13 |
accepted | Proof about that operator is self-adjoint |
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May 13 |
comment |
Proof about that operator is self-adjoint By expanding T(𝔼x) do you mean to divide Ex based on the new base? |
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May 13 |
comment |
Proof about that operator is self-adjoint how does it take this assumption into account though ? <T(Ex), Ey> = <Ex, T(Ey)> |
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May 13 |
comment |
Proof about that operator is self-adjoint You are right, I'm sorry, in my mind the line I was referring to what the most important and problematic part of the text. I've been struggling with it for the last 4 hours, and was more or less the last line I've paid attention to in the text. Sorry for the inconvenience. |
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May 13 |
comment |
Proof about that operator is self-adjoint sorry about that, meant "A" linear transformation, Updated the question though. thanks! |
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May 13 |
comment |
Proof about that operator is self-adjoint Hi fgp, I've edited my question as it wasn't clear enough. I was referring to another line I didn't quite understand Why is this true: If <T(Ex), Ey> = <Ex, T(Ey)> ==> Xt(BtA)y = Xt(AB)y |
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May 13 |
revised |
Proof about that operator is self-adjoint added 120 characters in body |
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May 13 |
asked | Proof about that operator is self-adjoint |
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May 6 |
asked | How can I develop a reduction formula for $\int \sin^n d x$ in 1 step jumps |
