Prove that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$= $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$

Question

QuestionProve that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$= $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$

My Approach R.H.S

$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$=$\frac{x^{2^{k-1}\left(1-x^{2}\right)}}{\left(1-x^{2^{k}}\right)\left(1-x^{2^{k-1}}\right)}$

i just don't know what else i can do?

Your step is not quite correct; note that $(x^{2^{k - 1}})^2 = x^{2^k}$. — theyaoster, Jan 10 '18 at 06:39

Rohan Shinde · Accepted Answer · 2018-01-10T06:53:07.667

5

Consider $x^{2^k}=m$

The equation reduces to $$\frac {\sqrt m}{1-m}=\frac {1+\sqrt m -1}{1-m}$$ $$=\frac {1+\sqrt m}{1-m}-\frac {1}{1-m}$$ $$=\frac {1}{1-\sqrt m} - \frac {1}{1-m}$$ Resubstitute value of $m$ to obtain desired proof

edited Jan 10 '18 at 06:53

answered Jan 10 '18 at 06:37

Rohan Shinde

9,737

Standard trick, pretty useful ! – jonsno Jan 10 '18 at 06:40
Glad you liked it – Rohan Shinde Jan 10 '18 at 06:40

score 2 · Answer 2 · answered Jan 10 '18 at 06:44

2

$$\begin{align} \frac{1}{1-x^{2^{k-1}}} - \frac{1}{1-x^{2^k}} &= \frac{(1-x^{2^k}) - (1-x^{2^{k-1}})}{(1-x^{2^{k-1}})(1-x^{2^k})}\\ &= \frac{x^{2^{k-1}}(1-x^{2^{k-1}})}{(1-x^{2^{k-1}})(1-x^{2^k})}\\ &= \frac{x^{2^{k-1}}}{1-x^{2^k}} \end{align}$$

answered Jan 10 '18 at 06:44

adfriedman

3,641

sirous · Answer 3 · 2018-01-10T11:17:19.993

2

$x^{2^k}=x^{2\times2^{k-1}}=(x^{2^{k-1}})^2$

⇒$1-x^{2^k}=(1-x^{2^{k-1}})(1+x^{2^{k-1}})$

⇒ $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^k}}=\frac{1+x^{2^{k-1}}-1}{1-x^{2^k}}=\frac{x^{2^{k-1}}}{1-x^{2^k}}$

edited Jan 10 '18 at 11:17

answered Jan 10 '18 at 06:51

sirous

10,751

Prove that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$= $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$

3 Answers3