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anonymous

  • one year ago

help please

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  1. anonymous
    • one year ago
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    find the dual basis of {(1,0,0),(0,1,0),(0,0,1) and explain

  2. anonymous
    • one year ago
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    @zzr0ck3r

  3. anonymous
    • one year ago
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    @oldrin.bataku

  4. anonymous
    • one year ago
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    its a table

  5. anonymous
    • one year ago
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    please continue

  6. anonymous
    • one year ago
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    notice that the 1 shifts each time.

  7. anonymous
    • one year ago
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    yes

  8. anonymous
    • one year ago
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    1, 0, 0 0,1,0 0,0,1 thats binary code

  9. anonymous
    • one year ago
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    yes

  10. anonymous
    • one year ago
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    the first column is 2^0 second column is 2^1 third column is 2 ^2

  11. anonymous
    • one year ago
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    how?

  12. anonymous
    • one year ago
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    you know how to count in tens. you know the first column is 0-9 the second column is 10 - 19 etc., remember the abacus? it is similar for the binary system

  13. anonymous
    • one year ago
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    ok

  14. anonymous
    • one year ago
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    I'll draw you a picture!

  15. anonymous
    • one year ago
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    |dw:1439037967941:dw|

  16. anonymous
    • one year ago
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    i don,'t still get why 2^ of numbers here.

  17. anonymous
    • one year ago
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    another picture in base 10 (decimal system, which you are more used to) |dw:1439038148182:dw|

  18. anonymous
    • one year ago
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    There are only 2 numbers 1 and zero that is why it is base 2 or 2^

  19. anonymous
    • one year ago
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    ok

  20. anonymous
    • one year ago
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    so how come the numbers they raised to? 2^n

  21. anonymous
    • one year ago
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    look up positional number systems in wikipedia it should explain it more to you

  22. anonymous
    • one year ago
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    The value that a digit taks on depends on its position, as I have shown

  23. anonymous
    • one year ago
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    ok. got it

  24. anonymous
    • one year ago
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    for example, 110 in decimal system is 1 x 100 + 1 x 10 + 1 x 1

  25. anonymous
    • one year ago
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    so, you did for base 2 but why base 10?

  26. anonymous
    • one year ago
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    I just showed you the same principle applies in base 2 as it applies in base 10 (decimal) which you were more familiar with

  27. anonymous
    • one year ago
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    but we are not using base 10 right since you said that only numbers are involved . ?

  28. anonymous
    • one year ago
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    the value of the number (e.g. 001) depends on itsa place or position) is what I am saying

  29. anonymous
    • one year ago
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    ok

  30. anonymous
    • one year ago
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    So of 1 is placed in the units column in the decimal system then it can only take the value between 0 and 9

  31. anonymous
    • one year ago
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    your question was about the dual basis of this, I have offered an answer in relation to that. i.e that the digits given are base 2

  32. anonymous
    • one year ago
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    so the answer is 7?

  33. anonymous
    • one year ago
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    No. It said find the dual basis of this, and explain. you have found the dual basis. reproduce my table i n relation to 2^0, "^1 etc, and then state that the value of oo1,010 depends on their position.

  34. anonymous
    • one year ago
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    you get 7 if you add the numbers

  35. anonymous
    • one year ago
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    ok, so the table with the position is the dual basis?

  36. anonymous
    • one year ago
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    for something to have a "dual basis" it must have two possible outcomes.i.e. binary. you show that as follows: |dw:1439039911382:dw|

  37. anonymous
    • one year ago
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    you will get 100% trust me

  38. anonymous
    • one year ago
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    now you are asked to explain: so here goes:( the value of the number zero or 1 depends on the position) that's it!

  39. anonymous
    • one year ago
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    ok thanks . i have a question on inner produte. can i ask?

  40. anonymous
    • one year ago
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    can i ask?

  41. anonymous
    • one year ago
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    compute the I2 inner product of x and y where x=1,1/2,1,3,1/4 and y=1/3,1/4,1/5

  42. anonymous
    • one year ago
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    that's not what the dual basis is. if you have a real inner product space \(V,\langle\cdot,\cdot\rangle\) then an ordered basis \(\{b_i\}\) induces a basis for the dual space \(V^*\) of linear maps \(V\to R\), such that $$b^i(v)=\langle b_i,v\rangle$$

  43. anonymous
    • one year ago
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    this is equivalent to the idea of the transpose, where we identify column vectors with \(v\in V\) and so row vectors with \(u\in V^*\)

  44. anonymous
    • one year ago
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    so what is the duel basis of my question sir?

  45. anonymous
    • one year ago
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    in other words, for an orthonormal basis \(\{e_i\}\), like you have, our dual basis looks like \(e^i\), so that \(e^i(e_i)=1\) and \(e^i(e_j)=0\) for \(i\ne j\). in other words, our dual basis lets us pick out things parallel to the corresponding original basis vectors

  46. anonymous
    • one year ago
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    so a dual basis for \(\{(1,0,0),(0,1,0),(0,0,1)\}\) looks like \(\{\delta_{1j},\delta_{2j},\delta_{3j}\}\) where \(\delta_{ij}\) denotes the Kronecker delta: $$\delta_{ij}=\left\{\begin{matrix}1&\text{if }i=j\\0&\text{otherwise}\end{matrix}\right.$$

  47. anonymous
    • one year ago
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    is that all?

  48. anonymous
    • one year ago
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    now in \(\ell^2(\mathbb N)\) we're talking about a sequence space and we have an inner product like: $$\langle x,y\rangle=\sum_{k=0}^\infty x_ky_k$$ for real sequences \(x,y\)

  49. anonymous
    • one year ago
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    $$\langle\left(1,\frac12,1,3,\frac14,0,0,0,\dots\right),\left(\frac13,\frac14,\frac15,0,0,0,\dots\right)\rangle=1\cdot\frac13+\frac12\cdot\frac14+1\cdot\frac15=\dots$$

  50. anonymous
    • one year ago
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    can you compute \(1/3+1/8+1/5\)?

  51. anonymous
    • one year ago
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    (40+15+24)/120=79/120

  52. anonymous
    • one year ago
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    @Michele_Laino

  53. anonymous
    • one year ago
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    @Loser66

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