Some (irrelevant to this thread) calculations produced these two expressions:
√1201+√(70+2*√1201)
&
√(35+2*√6)+√(1236-2*√6+2*√(42035-2402*√6))
which should be equal. I have checked with the Longfloat Lib on the HP 49G, & the two are equal to many decimal positions, but are they in fact exactly equal?
Assistance appreciated.
(Title edited to include HP Prime, 13:33, 26/6)
(06-26-2014 10:25 AM)Gerald H Wrote: [ -> ]are they in fact exactly equal?
True
(06-26-2014 10:45 AM)Thomas Klemm Wrote: [ -> ] (06-26-2014 10:25 AM)Gerald H Wrote: [ -> ]are they in fact exactly equal?
True
Thank you.
I believe the result is correct & dislike relying on the authority of some cloud-computing; Why should I trust Wolframalpha if I have no means of checking the result?
My version of Maple is antiquated & can't deal with the question, nor can the HP 49G - & even if they did return an intelligible answer, I'd still want to know how.
(06-26-2014 10:25 AM)Gerald H Wrote: [ -> ]<clipped>these two expressions:
√1201+√(70+2*√1201)
&
√(35+2*√6)+√(1236-2*√6+2*√(42035-2402*√6))
<clipped>but are they in fact exactly equal?
Assistance appreciated.
50G says no
(06-26-2014 11:01 AM)CosmicTruth Wrote: [ -> ] (06-26-2014 10:25 AM)Gerald H Wrote: [ -> ]<clipped>these two expressions:
√1201+√(70+2*√1201)
&
√(35+2*√6)+√(1236-2*√6+2*√(42035-2402*√6))
<clipped>but are they in fact exactly equal?
Assistance appreciated.
50G says no
Thank you.
So now we have two authorities dsagreeing(see post #2)?
I have now tried the expressions on HP Prime CAS: for == a zero is returned & for - a value of -2.27373675443E-13.
The numerical value for - is certainly wrong.
My 2 cents:
I ran it in the newRPL demo at 2007 digits precision, and the difference between both expressions came out 1e-2005, so I'd say they are equal at least up to the first 2000 digits.
Claudio
To get an algebraic proof, I don't have the time but I think the key is:
Code:
1201 = 35^2 - 2^2*6
Claudio
(06-26-2014 04:08 PM)Claudio L. Wrote: [ -> ]My 2 cents:
I ran it in the newRPL demo at 2007 digits precision, and the difference between both expressions came out 1e-2005, so I'd say they are equal at least up to the first 2000 digits.
Claudio
Thank you for the confirmation - I hadn't tested to such precision.
The means by which the two expressions arose implies, I believe, equality & I'm not bright enough to demonstrate this equality.
More precision will (hopefully) corroborate equality, but a convincing reasoning would settle the matter.
OK, first let's notice that 1201 is prime, 42035=35*1201 and 1236=35+1201. Now rewrite the longer expression:
\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{1201+35-2\sqrt{6}+2\sqrt{\left(35-2\sqrt{6}\right)1201}}
\end{equation}
That is the square of a sum:\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{\left(\sqrt{1201}+\sqrt{35-2\sqrt{6}}\right)^2}
\end{equation}
You don't need to worry about the absolute value, it's simply:\begin{equation}\sqrt{1201}+\sqrt{35+2\sqrt{6}}+\sqrt{35-2\sqrt{6}}
\end{equation}
If a>b it's trivial to prove that:\begin{equation}\sqrt{a+b}+\sqrt{a-b}=\sqrt{2a+2\sqrt{a^2-b^2}}\end{equation}
In this case: \begin{equation}\sqrt{70+2\sqrt{1225-4\cdot 6}}=\sqrt{70+2\sqrt{1201}}\end{equation}
There you go.
(You guys should use paper and pencil more often
)
This reminds me of Dedekind's Theorem that \(\sqrt{2} \sqrt{3} = \sqrt{6}\). A very readable account is in the article "Dedekind's Theorem: ..." by Fowler in The American Mathematical Monthly Vol. 99, No. 8, Oct., 1992, p.725.
(06-26-2014 05:58 PM)Manolo Sobrino Wrote: [ -> ]OK, first let's notice that 1201 is prime, 42035=35*1201 and 1236=35+1201. Now rewrite the longer expression:
\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{1201+35-2\sqrt{6}+2\sqrt{\left(35-2\sqrt{6}\right)1201}}
\end{equation}
That is the square of a sum:\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{\left(\sqrt{1201}+\sqrt{35-2\sqrt{6}}\right)^2}
\end{equation}
You don't need to worry about the absolute value, it's simply:\begin{equation}\sqrt{1201}+\sqrt{35+2\sqrt{6}}+\sqrt{35-2\sqrt{6}}
\end{equation}
If a>b it's trivial to prove that:\begin{equation}\sqrt{a+b}+\sqrt{a-b}=\sqrt{2a+2\sqrt{a^2-b^2}}\end{equation}
In this case: \begin{equation}\sqrt{70+2\sqrt{1225-4\cdot 6}}=\sqrt{70+\sqrt{1201}}\end{equation}
There you go.
(You guys should use paper and pencil more often )
The last calculation line is a typo?
(06-26-2014 06:51 PM)Gerald H Wrote: [ -> ]The last calculation line is a typo?
Of course it was, fixed now. (LaTeX here seems to cause me dyslexia
)
I take my hat off to you,Manolo Sobrino. Bravo!
HP50G calculator for sale or trade for good pencil and paper pad.
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