perplexus.info :: Just Math : Complex

Complex (Posted on 2018-12-22)

A = 1 + x³/3! + x⁶/6! + x⁹/9! + ...
B = x + x⁴/4! + x⁷/7! + x¹⁰/10! + ...
C = x²/2! + x⁵/5! + x⁸/8! + x¹¹/11! + ...

Find the value of A³ + B³ + C³ - 3ABC.

No Solution Yet Submitted by Danish Ahmed Khan

Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

A start

| Comment 1 of 3

A + B + C = taylor series for e^x,

so A + B + C = e^x

Then e^3x = (A + B + C)^3

expanding and rearranging gives

A^3 + B^3 + C^3 - 3ABC

= e^3x -3(AB + AC +BC)(A+B+C)

= e^3x -3(AB + AC + BC)e^x

That's as far as I got.

Posted by Steve Herman on 2018-12-23 18:10:40

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info