Real formula?

lemonrealized

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Jan 29, 2018
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Hi,

I have added a image. It is this a real formula?
 

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tkhunny

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What does your question even mean? Is there a formula test for REALness?

Just read it careful for what t claims.
 

mmm4444bot

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Oct 6, 2005
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I'm not sure why you're asking.

Are you, perhaps, a beginning algebra student, and you're wondering if the statements in that image are valid or faked? (You posted on the Beginning Algebra board).

Or, you are not a beginning algebra student, and you posted on the wrong board, maybe?

What you posted is not beginning algebra, and it looks like valid math. It seems to conclude that -- in some sequence of numbers -- the difference between adjacent elements is non-negative. That could also be interpreted as a conclusion that the number sequence is decreasing. :cool:
 

JeffM

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Sep 14, 2012
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Like mmm, I do not exactly what you are asking or why it is posted in beginners' algebra.

If the only information given is what is shown in the excerpt posted, that excerpt is logically flawed.

\(\displaystyle x_{k+1}^2 = \dfrac{1}{4} * \left (x_k^2 + 2A + \dfrac{A^2}{x_k^2} \right ) = \dfrac{x_k^2 + 2A + \dfrac{A^2}{x_k^2}}{4} \implies\)

\(\displaystyle x_{k+1}^2 - A = \dfrac{x_k^2 + 2A + \dfrac{A^2}{x_k^2}}{4} - A = \dfrac{x_k^2 + 2A + \dfrac{A^2}{x_k^2}}{4} - \dfrac{4A}{4} \implies\)

\(\displaystyle x_{k+1}^2 - A = \dfrac{x_k^2 - 2A + \dfrac{A^2}{x_k^2}}{4} - A = \left ( \dfrac{x_k - \dfrac{A}{x_k}}{2} \right )^2 \ge 0 \implies x_{k+1}^2 - A \ge 0 \implies\)

\(\displaystyle x_{k+1}^2 \ge A \implies \sqrt{x_{k+1}^2} \ge \sqrt{A} \implies |x_{k+1}| \ge \sqrt{A} \not \implies x_{k + 1} \ge \sqrt{A}.\)

Moreover the indexing is a bit weird in the final conclusion. Does it relate to k = 1 or not?
 
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