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Sum of a sequence, Sum of lengths of connected straight lines perpendicular to rotated lines at a certain angle

ÀÛ¼ºÀÚ Uploader : billy ÀÛ¼ºÀÏ Upload Date: 2019-11-19º¯°æÀÏ Update Date: 2020-02-19Á¶È¸¼ö View : 1703

Sum of the lengths of the connected lines that are perpendicular to lines rotated at a constant angle around the original point.

If the length of the line OA1 is L, the length of line OA2 is L * cos(¥è) and a the length of the line A1A2 is L * sin(¥è). Using this relationship, we get :

With n=1, L : L*cos(¥è) : L*sin(¥è)
With n=2, L*cos(¥è) : L*(cos(¥è))^2 : L*cos(¥è)*sin(¥è)
With n=3, L*(cos(¥è))^2 : L*(cos(¥è))^3 : L*(cos(¥è))^2*sin(¥è)
With n=4, L*(cos(¥è))^3 : L*(cos(¥è))^4 : L*(cos(¥è))^3*sin(¥è)
  ¡¦
With n=k, L*(cos(¥è))^(k-1) : L*(cos(¥è))^k : L*(cos(¥è))^(k-1)*sin(¥è)

Then, the sum of lengths of lines is :

Sum = A1A2 + A2A3 + A3A4 + ¡¦

  = L*sin(¥è)+L*cos(¥è)*sin(¥è)+L*(cos(¥è))^2*sin(¥è)+L*(cos(¥è))^3*sin(¥è)+¡¦

This is equal to the sum of a geometric sequence of which the first term is L*sin(¥è) and the common ratio is cos(¥è).

From Sum = L*sin(¥è)*(1-(cos(¥è))^n)/(1-cos(¥è)), if n ¡æ ¡Ä, (cos(¥è))^n = 0, then,

Sum = L*sin(¥è)/(1-cos(¥è))


*** Âü°í¹®Çå[References] ***

Sum = L*sin(¥è)/(1-cos(¥è))
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