Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $

Question

Please help me find the sum

$$ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $$

Answer 1

$$ \frac1{x-y} + \frac1{x+y} = \frac{2x}{x^2-y^2}$$

Proceeding in this fashion we would be left with

$$ \text{The sum } = \frac{16x^{15}}{x^{16}-y^{16}} + \frac{16x^{15}}{x^{16}+y^{16}} = \frac{32x^{31}}{x^{32}-y^{32}} $$

Answer 2

Note that each term is of the form $$\dfrac{df_k/dx}{f_k} = \dfrac{d(\log(f_k)))}{dx}$$ Hence, $$\sum_{k=0}^n \dfrac{df_k/dx}{f} = \sum_{k=1}^n \dfrac{d(\log(f_k)))}{dx} = \dfrac{d(\sum_{k=0}^n \log(f_k)))}{dx} = \dfrac{d(\log(f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_n))}{dx}$$In your case, $f_0 f_1 f_2 \cdots f_n$ reduces to give a nice short expression, which can be easily differentiated. Move the cursor over the gray area for a complete answer.

In your case, $f_0 = (x-y)$, $f_n = x^{2^{n-1}} + y^{2^{n-1}}$ for $n > 0$. Hence, $$f_0 f_1 f_2 \cdots f_n = (x-y)(x+y)(x^2+y^2)\cdots (x^{2^{n-1}} + y^{2^{n-1}}) = x^{2^n} - y^{2^n}$$ Hence, $$\dfrac{d(\log(f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_n))}{dx} = \dfrac{2^n x^{2^n-1}}{x^{2^n}-y^{2^n}}$$ In your case, $n=5$.

Answer 3

Adding the terms together, you should get: $$ \sum_{\text{all terms}} = \frac{(32 x^{31})}{(x^{32}-y^{32})} $$ This result is obtained by using the LCD, (Least Common Denominator).

LCD: $$ (x-y) (x+y) (x^2+y^2) (x^4+y^4) (x^8+y^8) (x^{16}+y^{16}) =\\ (x^2-y^2)(x^2+y^2)(x^4+y^4) (x^8+y^8) (x^{16}+y^{16})=\\ (x^4-y^4)(x^4+y^4)(x^8+y^8) (x^{16}+y^{16})=\\ (x^8-y^8)(x^8+y^8)(x^{16}+y^{16})=\\ (x^{16}-y^{16})(x^{16}+y^{16})=\\ (x^{32}-y^{32}) =\\ \text{LCD(LCD(LCD(LCD(LCD( $x-y$, $x+y$), $x^2+y^2$), $x^4+y^4$), $x^8 + y^8$), $x^{16}+y^{16}$)} $$

Note

It would be preferable to move from left to right simplifying the pairwise additions. I included the LCD so you could see the general pattern in the denominators as you progress in your simplification from left to right.

Answer 4

They're arranged rather nicely. Proceed from left to right and keep using the formula $(a-b)(a+b)=a^2-b^2$ to rewrite the two terms you're about to add with a common (unfactored) denominator. You'll get a good deal of additive cancellation in the numerators, so it won't be all that messy at any stage.

Answer 5

$$ \left(\left(\left(\left(\left(\frac{1}{x-y}+\frac{1}{x+y}\right)+\frac{2x}{x^2+y^2}\right)+\frac{4x^3}{x^4+y^4}\right)+\frac{8x^7}{x^8+y^8}\right)+\frac{16x^{15}}{x^{16}+y^{16}}\right) $$