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amistre64

  • 3 years ago

Think of 3 rules, such that they generate 2,3,6 as the first 3 terms of a sequence.

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  1. amistre64
    • 3 years ago
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    1/2 3/2 -4 i came up with this one:\[\frac{1}{2}\left(\frac{n-1.5}{|n-1.5|}+\frac{3n-7.5}{|n-2.5|}-\frac{8n-280}{|n-3.5|}\right)\]

  2. amistre64
    • 3 years ago
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    lol, i forgot to delete my reference scribbles

  3. amistre64
    • 3 years ago
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    i was also considering coming up with something such that:\[\int_{0}^{n}f(x)dx=2,3,6~for~n=1,2,3\]

  4. amistre64
    • 3 years ago
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    the first one i thought of was\[a_n=n^2-2n+3\]

  5. amistre64
    • 3 years ago
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    |dw:1346159138909:dw| hmm; such that the first area is 2, and the second is 1, and the third is 3

  6. amistre64
    • 3 years ago
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    if i could determine the slopes in that figure, i could integrate the function to get what im thinking of :)

  7. amistre64
    • 3 years ago
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    |dw:1346159507196:dw|b-a=1\[\int_{a}^{b}k~dx=F(b)-F(a)=1\]F(b)=4, F(a)=-3

  8. amistre64
    • 3 years ago
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    err, F(a) = 3

  9. amistre64
    • 3 years ago
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    |dw:1346159801695:dw| 0 to b , b<1 ... its like right on the tip of my brain

  10. amistre64
    • 3 years ago
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    |dw:1346160036988:dw| at a, A=2, at (b-a)/2, A=0 so there has to be a sweet spot between a and the midpoint that satisfies my quandry

  11. amistre64
    • 3 years ago
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    if b= 3/4; and a = 1/4 \[\frac{4}{3/4}x=\frac{16}{3}x; \int\to\ \frac83 \left((\frac34)^2-(\frac14)^2\right)\]\[\frac83 \left(\frac{9-1}{16}\right)=1\frac13\] so its between 1/4 and 1/2 :)

  12. amistre64
    • 3 years ago
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    |dw:1346161272353:dw| hmmm

  13. amistre64
    • 3 years ago
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    not B=8, 4+4=8/2=4; that triangle area there is B=4

  14. amistre64
    • 3 years ago
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    if i guess that: (4+d)/2 = 3; d=2 mx-d 4+2 = 6, 6 in 1 = 6x, 6x-2 3x^2-2x, (0,1) = 3-2=1

  15. amistre64
    • 3 years ago
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    |dw:1346162349632:dw|

  16. amistre64
    • 3 years ago
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    2,3,6 is my goal; and i wonder if alternating signs, and doubling would give me my needed slopes, making the next one 12 12x-2; 6x^2-2x = 4, not 3 it was a nice guess

  17. amistre64
    • 3 years ago
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    int mx-2 , [0,1] m/2 - 2 = 3 m/2 = 5 m=10

  18. amistre64
    • 3 years ago
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    |dw:1346163082261:dw|

  19. amistre64
    • 3 years ago
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    -a-b-c = 4 a-b-c=-6 a+b-c=10 rref{{-1,-1,-1,4},{1,-1,-1,-6},{1,1,-1,10}} -5,8,-7 f(x) = -5|x-1|+8|x-2|-7|x-3| and since n=0 = -10; lets add 10 to it to move it to the proper spot \[\int_{0}^{n}f(x) = -5|x-1|+8|x-2|-7|x-3|+10~dx=\{2,3,6,...\}~:~n=1,2,3,...\]

  20. amistre64
    • 3 years ago
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    \[\int_{0}^{n}f(x)~dx\]\[\int_{0}^{n} (-5|x-1|+8|x-2|-7|x-3|+10)~dx=\{2,3,6,...\}~:~n=1,2,3,...\] that fixes the typos

  21. amistre64
    • 3 years ago
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    ima goofy goober!!! lol

  22. anonymous
    • 3 years ago
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    wow!

  23. amistre64
    • 3 years ago
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    i know right, i should be soo ashamed

  24. anonymous
    • 3 years ago
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    i guess my idea of a second degree polynomial is too prosaic. you probably did that first anyway

  25. anonymous
    • 3 years ago
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    maybe change it to "find the most arcane sequences whose first three terms are ..."

  26. amistre64
    • 3 years ago
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    i did .... and my second original idea was:\[a_{n+1}=a{n}*a_{n-1}~:~a_1=2, a_2=3\]

  27. anonymous
    • 3 years ago
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    that is a good one

  28. amistre64
    • 3 years ago
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    if by arcane you mean deviously clever ... i agree

  29. anonymous
    • 3 years ago
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    i really meant "off the wall" not arcane

  30. amistre64
    • 3 years ago
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    i wonder what the math teachers gonna give me as a grade on the homework this year; or if shell even read it

  31. anonymous
    • 3 years ago
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    you will get a gold star for cleverness

  32. amistre64
    • 3 years ago
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    i gotta hobble over to physics and 2 other classes for awhile; yall try not to break stuff while im gone

  33. UnkleRhaukus
    • 3 years ago
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    \[\large{s_n=2n-\frac n4\left(1+(-1)^n\right)}\] \[s_1=2-\frac 14\left(1+(-1)^1\right)=2-0=2\] \[s_2=2\times2-\frac 24\left(1+(-1)^2\right)=4-\frac12(2)=3\] \[s_3=2\times3-\frac 34\left(1+(-1)^3\right)=6-0=6\]

  34. anonymous
    • 3 years ago
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    \[a_n=(-1)^{n}n^2\cos(\pi n)+(-1)^{n+1}2\cos(\pi n)+(-1)^n3\cos(\pi n)\]

  35. eliassaab
    • 3 years ago
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    Why is interesting? I do not know. Here is my simple contribution. \[ a_1=2\\ a_2=3\\ a_n=a_{n-1}a_{n-2},\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=a_{n-1}+a_{n-2}+1,\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=a_{n-1}-a_{n-2}+5,\quad n>2 \]

  36. eliassaab
    • 3 years ago
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    Here are two more. You can have infinite such rules. \[ a_1=2\\ a_2=3\\ a_n=2 a_{n-1} ,\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=3 a_{n-2} ,\quad n>2 \]

  37. amistre64
    • 3 years ago
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    the question was so bland that i figured id see if there were more interesting ways to develop a sequence that began with the given digits.

  38. amistre64
    • 3 years ago
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    if my teacher insists that i have to do the homework for a grade .... im at least going to try to get some fun out of it :)

  39. Herp_Derp
    • 3 years ago
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    \[a_n=\Pi (n)-\Gamma (n) +2\] \(\Pi (n)=n!\) and \(\Gamma (n) =(n-1)!\) but it just looks so much cooler that way.

  40. amistre64
    • 3 years ago
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    lol, awesome

  41. Herp_Derp
    • 3 years ago
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    This one is just... idk \[\large a_n=\left\lfloor\sqrt[n]{n+\Big\lfloor e^{\frac{1}{2}n^2}\Big\rfloor}\right\rfloor +\left\lfloor\int_0^n \frac{\frac{1}{2}xe^{-\sqrt{x}}}{\sin (x)}\,\text{d}x\right\rfloor\]

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