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- Thread starter mesa
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AlephZero

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Or you could do something like the series expansion of ##\frac 1 5 \sin \pi##.

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Excellent! We can get those by changing the initial value and inputting different integers into the series.You can easily invent a geometric series that sums to any value you want.

Or you could do someting like the series expansion of ##\frac 1 5 \sin \pi##.

Do you know of any other ways to get to these values outside of geometric and trigonometric series? For example do we have a factorial series for calculating these values?

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Office_Shredder

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That won't get him 1/5, though. :tongue:Or you could do something like the series expansion of ##\frac 1 5 \sin \pi##.

I'm sure that was a typo. mesa, he probably meant ##\frac{\sin\frac{\pi}{2}}{5}##.

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jgens

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Indeed.

mesa, you should look back at your other thread on infinite series. Many of the same series (plural) apply here.

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Borek

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Get any sum that yields any finite number, multiply it by a "normalizing factor".

[tex]e=\sum_0^\infty\frac 1 {n!}[/tex]

so

[tex]\frac 1 5 = \sum_0^\infty\frac 1 {5en!}[/tex]

[tex]e=\sum_0^\infty\frac 1 {n!}[/tex]

so

[tex]\frac 1 5 = \sum_0^\infty\frac 1 {5en!}[/tex]

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http://atsol.fis.ucv.cl/dariop/sites...ij_Engl._2.pdf

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To get an idea of what is out there. Do we have a 'general solution' outside of geometric series, trig, and normalizing factors for solving all whole integer fractions with '1' in the numerator through infinite series?

Very good, it can be tough to spot errors late on New Years Eve... :)That won't get him 1/5, though. :tongue:

I'm sure that was a typo. mesa, he probably meant ##\frac{\sin\frac{\pi}{2}}{5}##.

Hah!

Yes they do, I was curious if we had more 'general' solutions for infinite series that would give whole integer fractions.Indeed.

mesa, you should look back at your other thread on infinite series. Many of the same series (plural) apply here.

That looks familiar :)Get any sum that yields any finite number, multiply it by a "normalizing factor".

[tex]e=\sum_0^\infty\frac 1 {n!}[/tex]

so

[tex]\frac 1 5 = \sum_0^\infty\frac 1 {5en!}[/tex]

Hmmm, that link doesn't seem to be working?

http://atsol.fis.ucv.cl/dariop/sites...ij_Engl._2.pdf

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AlephZero

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More of a brain-fart than a typo. I divided ##2\pi## by ##2## and got ##\pi##. Full marks for arithmetic, but zero marks for doing the wrong sum.I'm sure that was a typo.

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1/5 = 0.19999...

1/7 = 0.142857142857...

1/7 = 0.142857142857...

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Nah. We all make mistakes or have lapses in genius from time to time. It's all good.More of a brain-fart than a typo. I divided ##2\pi## by ##2## and got ##\pi##. Full marks for arithmetic, but zero marks for doing the wrong sum.

What do you mean "more general"?Yes they do, I was curious if we had more 'general' solutions for infinite series that would give whole integer fractions.

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I should probably have said, 'general solution' for an infinite series that can be used to get any whole integer fraction by changing the index of summation. For example, an infinite series that would be able to compute 1/5 and 1/7 simply by changing the index. Do you know if we have anything like this?What do you mean "more general"?

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That certainly would work much like Boreks suggestion of using a 'normalizing factor'.

Do you know of any series that will equal

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Consider the sequence $$a_i=\{1,-1,\frac{1}{2},\frac{1}{2},-1,\frac{1}{3},\frac{1}{3},\frac{1}{3},-1,...\}.$$That certainly would work much like Boreks suggestion of using a 'normalizing factor'.

Do you know of any series that will equalallwhole integer fractions by simply changing the lower limit of summation?

##\displaystyle \sum_{n_0\leq i\leq n}a_i## should cover everything you want. :tongue:

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Brilliant!Consider the sequence $$a_i=\{1,-1,\frac{1}{2},\frac{1}{2},-1,\frac{1}{3},\frac{1}{3},\frac{1}{3},-1,...\}.$$

##\displaystyle \sum_{n_0\leq i\leq n}a_i## should cover everything you want. :tongue:

...and I need to brush up on my writing comprehension skills, you guys are too clever

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