An Extrapolation of a Beautiful Series

Calculus Level 3

( x + y ) ( x − y ) + 1 2 ! ( x + y ) ( x − y ) ( x 2 + y 2 ) + 1 3 ! ( x + y ) ( x − y ) ( x 4 + y 4 + x 2 y 2 ) + … = ? \begin{aligned} && (x+y)(x-y) \\ &&+ \dfrac{1}{2!} (x+y)(x-y)(x^2 + y^2) \\ &&+ \dfrac{1}{3!} (x+y)(x-y)(x^4 + y^4 + x^2 y^2 ) \\ &&+ \ldots \\ &&= \ ? \end{aligned}

Try my set
e x 2 + e y 2 e^{x^2} + e^{y^2} e x 2 − e y 2 e^{x^2} - e^{y^2} None of these options e x 2 − y 2 e^{x^2 - y^2} e x 2 + y 2 e^{x^2 + y^2}

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Vishwak Srinivasan by Vishwak Srinivasan , 24, India

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1 solution

Chew-Seong Cheong
May 20, 2015

( x + y ) ( x − y ) + 1 2 ! ( x + y ) ( x − y ) ( x 2 + y 2 ) + 1 3 ! ( x + y ) ( x − y ) ( x 2 + y 2 + x 2 y 2 ) + . . . = x 2 − y 2 + 1 2 ! ( x 4 − y 4 ) + 1 3 ! ( x 6 − y 6 ) + . . . = ( 1 + x 2 + 1 2 ! x 4 + 1 3 ! x 6 + . . . ) − ( 1 + y 2 + 1 2 ! y 4 + 1 3 ! y 6 + . . . ) = e x 2 − e y 2 (x+y)(x-y)+\dfrac{1}{2!}(x+y)(x-y)(x^2+y^2)\\ \quad \quad +\dfrac{1}{3!}(x+y)(x-y)(x^2+y^2+x^2y^2)+... \\ = x^2-y^2 + \dfrac{1}{2!} (x^4-y^4) + \dfrac{1}{3!} (x^6-y^6) + ... \\ = \left(1 + x^2 + \dfrac{1}{2!} x^4 + \dfrac{1}{3!} x^6 + ...\right) - \left(1 + y^2 + \dfrac{1}{2!} y^4 + \dfrac{1}{3!} y^6 + ...\right) \\ = \boxed{e^{x^2} - e^{y^2}}

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Nice one sir.

Vishwak Srinivasan - 6 years ago

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Thank you. Set more problems for every one.

Chew-Seong Cheong - 6 years ago

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Sure will do sir.

Vishwak Srinivasan - 6 years ago

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