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Halley's
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You are currently reading a thread in /sci/ - Science & Math

Does this simplify further?
>>
Yes.
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>>7761421
Is that denominator: [eqn] \ln (2) \left ( 2^{2y_n} + x 2^{y_n} \right ) [/eqn] or [eqn] \ln \left ( 2 \left ( 2^{2y_n} + 2^{y_n} \right ) \right ) [/eqn]
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>>7761483
Yes
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>>7761483
[eqn] \ln (2) \left ( 2^{2y_n} + x 2^{y_n} \right ) [/eqn]
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>>7761483
Where do you see ln?
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>>7761421
You can cancel a 2^y_n
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>>7761809
denominator
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>>7761789
In that case factor out a 2^{y_n} in both the numerator and denominator and cancel.
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>>7761820
>>7761842
Ahhh, thanks for un-sticking me. Love when the answer has been staring me in the face the whole time.

So it boils down to:
y_{n+1}=y_n\ -\frac{2\left(2^{y_n}-x\right)}{\ln (2)\left(2^{y_n}+x\right)}

That form looks vaguely familiar to me, but maybe that's as far as I can get
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>>7762262
[eqn]y_{n+1}=y_n\ -\frac{2\left(2^{y_n}-x\right)}{\ln (2)\left(2^{y_n}+x\right)}[/eqn]
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>>7762270
[eqn] y_{n+1}=y_n\ -\frac{2\left(2^{y_n}-x\right)}{\ln (2)\left(2^{y_n}+x\right)} [/eqn]
omg. where can i get a plugin for LaTeX so this isn't so painful?
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>>7762262
>>7762270
>>7762275
\frac{} {} not \frac{}{} or else LaTeX just wont work on here

(from the first post)
$y_{n+1}=y_n - \frac{2 \left(2^{y_n}-x\right)} { \ln (2) \left(2^{y_n}+x \right)}$