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anonymous

  • 5 years ago

The polynomial 1-x+x ^{2}-x ^{3}+...-x ^{15}+x ^{16}-x ^{17} can be written as a polynomial in y=x+1. Find the coefficient of y ^{2}

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  1. amistre64
    • 5 years ago
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    i cant make head nor hair of it

  2. myininaya
    • 5 years ago
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    So I guess we should replace x with y-1

  3. myininaya
    • 5 years ago
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    \[1-(y-1)+(y-1)^2-(y-1)^3+(y-1)^4 \cdots \] we really don't need to right this any further all of them will be larger degree than y^2

  4. myininaya
    • 5 years ago
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    write*

  5. myininaya
    • 5 years ago
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    that is how i see it anyways

  6. Zarkon
    • 5 years ago
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    you could look at the Taylor polynomial

  7. amistre64
    • 5 years ago
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    doesnt (y-1)^3 expand with a y^2 in it?

  8. amistre64
    • 5 years ago
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    y^3, 3y^2, 3y, 1

  9. amistre64
    • 5 years ago
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    expanding all those terms would create y^2 s

  10. myininaya
    • 5 years ago
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    oh yeah you are right

  11. myininaya
    • 5 years ago
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    that wouldn't be efficient lol

  12. myininaya
    • 5 years ago
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    so we should look at zarkon's way

  13. amistre64
    • 5 years ago
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    well, if its the y^2 coeffs; its the pascal triangle diag then

  14. amistre64
    • 5 years ago
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    2,1,0 ^ 1 3 6 ...

  15. amistre64
    • 5 years ago
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    assuming myins interp is good :)

  16. amistre64
    • 5 years ago
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    ..... the L of the pascal .... thats equal to the next ones entry ....

  17. amistre64
    • 5 years ago
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    |dw:1327354588092:dw|

  18. Zarkon
    • 5 years ago
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    you want a polly that looks like this \[\sum_{k=0}^{17}a_ky^k\] there \(y=x+1\)

  19. Zarkon
    • 5 years ago
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    *where y=x+1

  20. amistre64
    • 5 years ago
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    1 1* 1 1 2* 1 1 3 3* 1 1 4 6 4* 1 1510 (10) 5 1

  21. amistre64
    • 5 years ago
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    but i think my idea has + and - signs to deal with still

  22. Zarkon
    • 5 years ago
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    I get 816 as the answer

  23. amistre64
    • 5 years ago
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    \[{{18}\choose {3}}=816.too\]

  24. asnaseer
    • 5 years ago
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    I notice that the person you guys are trying to assist has not said a word yet. @kwenisha - does all this make any sense to you? have you attempted it yourself using a different method? any thoughts?

  25. Zarkon
    • 5 years ago
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    using the Taylor polly amounts to computing \[\frac{\displaystyle\sum_{k=1}^{17}k(k-1)}{2}\]

  26. anonymous
    • 5 years ago
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    @Asnaseer, I am actually trying understand it while working with my partner... I apologize if I come off as just trying to get the answer and not trying to understand it...

  27. asnaseer
    • 5 years ago
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    thats fine @kwenisha - I was just worried in case the experts above left you dazed. :)

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