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It evaluates: (2/3)*(1/2)=1/3, so there is nothing special there. On the other hand, it's

Intergral square tea times. Three pie over nine equal log cubic root. Yeee...

Or something like that. Sounds like something from Alice in Wonderland.

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I got it...the poster before me is actually on the right track...

There was an Old Man with a beard,

Who said, 'It is just as I feared!

Two Owls and a Hen,

Four Larks and a Wren,

Have all built their nests in my beard!'

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I got it...the poster before me is actually on the right track...

There was an Old Man with a beard,

Who said, 'It is just as I feared!

Two Owls and a Hen,

Four Larks and a Wren,

Have all built their nests in my beard!'

Very good!! Steeleman23 knows the answer... The above hint will be sufficient enough to help most people... I'm proud you guys are getting it so quick :D

I will post the "exact" answer (and explanation) in a spoiler tomorrow (it's 12pm here)...

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I believe the significance of the equation is that

the equation links 3 (or 4) very famous people in mathematics in one equation, Newton (calculus), Ptolemy or Pythagoreas (trig) and Euler (the constant e).

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What I wanna know is...how long did it take whoever originally came up with it to figure it out? I mean it incorporates a lot of different types of math and it still fits the scheme.

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It evaluates: (2/3)*(1/2)=1/3, so there is nothing special there. On the other hand, it's

Intergral square tea times. Three pie over nine equal log cubic root. Yeee...

Or something like that. Sounds like something from Alice in Wonderland.

Sorry for being obtuse here... but how does the left-side of the equation evaluate to (2/3)*(1/2)??

The cosine of 3pi/9 is approximately 1 according to my calculator, and the integral of t^2 dt is t^3/3 which, as t approaches the cube root of 3 = 1/3. That looks like (1/3) * 1 to me. What am I missing here.

I know my math is pretty rusty, but am I just way off in the weeds?

Cheers,

Chris

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The cosine of 3pi/9 is approximately 1 according to my calculator

I had the same problem so don't feel bad. Make sure your calculator is in radian mode vs degree mode when you do the cosine. I saw like .998 for the cosine value and just assumed it was supposed to be 1 also and figured i didn't use enough significant figures. But I remembered the windows calculator uses degree mode as default. ;)

Also, the integral evaluates to 2/3....(t^3)/3 evaluated from 1 to the cube-root-of-3...simplifies to 1 - 1/3.

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I have the solution, and I now hate you. My friend (Computer Science Major) and I (Electrical Engineering Major) just spent an hour and a half trying to figure out the meaning. My friends girlfriend (English Major) poked her head in for 5 seconds and understood it. You're terrible.

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i hope random7 posts the answer soon i have no clue what imn doin

Ok... I think you've had enough... Here's the answer:

It's a Math Limerick... When read out aloud, it reads as follows:

The integral t-squared dt

From one to the cube root of three

Times the cosine

Of three pi over nine

Equals log of the cube root of e

For those that are not familiar with Limericks, they are short poems that have the form AABBA, such as above, or:

The limerick packs laughs anatomical

In space that is quite economical,

But the good ones I've seen

So seldom are clean,

And the clean ones so seldom are comical.

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What I wanna know is...how long did it take whoever originally came up with it to figure it out? I mean it incorporates a lot of different types of math and it still fits the scheme.

It's just like any other puzzle (creating Math Limericks)... I was going to set the challenge in this thread... But I think some of you are tired of this thread...

For example in this case, the composer started with "dt", "three"... "Cosine" and "nine"... And went from there... The reason for using "e" as the final line, is that the log of e to the power of anything, is that power... So it gives the composer the freedom to make the left hand exp​ression what ever he wanted... And he just has to raise the e to the answer to the left hand side... And to make it a nice equation... He used 3pi/9 in the cosine to evaluate to one half, if it were a sine in the Limerick, he would have used "three halves of pi on nine"... But it is seen that using cosine is neater and shorter (ie. fits the Limerick structure better)...

What I'm getting at is, the composer used logic and math to create it... And I believe you all have it in you to create your own... Here is one more for the road:

1cc1827236ed81aab83998b3290f05a1.png

A dozen, a gross, and a score

Plus three times the square root of four

Divided by seven

Plus five times eleven

Is nine squared and not a bit more

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I had the same problem so don't feel bad. Make sure your calculator is in radian mode vs degree mode when you do the cosine. I saw like .998 for the cosine value and just assumed it was supposed to be 1 also and figured i didn't use enough significant figures. But I remembered the windows calculator uses degree mode as default. ;)

Also, the integral evaluates to 2/3....(t^3)/3 evaluated from 1 to the cube-root-of-3...simplifies to 1 - 1/3.

Thanks, Steeleman!

I totally spaced on the radian thing as well... As for the integral, this is my rusty calculus brain... when integrating you take the difference of the limits? I think I vaguely remember that...

Cheers,

Chris

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This is actually something from Wikipedia, so it's not my own: What is special about this mathematical expression?

3d0d69d251554fad1c3ad6b36b5d9a78.png

I'll give it a few day's before posting a spoiler (or the straight answer). Hope you all have fun with this one...

Ok... so after my completely erroneous pedestrian guess, and with great help from those who have come before, I will give it a go:

The integral of t-squared dt,

From 1 to the cube-root of 3,

Times the cosine

of 3 pi over 9

Equals ln the cube-root of e

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Sorry for being obtuse here... but how does the left-side of the equation evaluate to (2/3)*(1/2)??

The cosine of 3pi/9 is approximately 1 according to my calculator, and the integral of t^2 dt is t^3/3 which, as t approaches the cube root of 3 = 1/3. That looks like (1/3) * 1 to me. What am I missing here.

I know my math is pretty rusty, but am I just way off in the weeds?

Cheers,

Chris

I studied my math back when they did not come up with calculators yet.

So the integral of t**2 is (1/3)* t**3. If you plug in the boundaries and subtract lower from upper, you get 2/3. Cosine of 60 degrees is 1/2, and I guess you have figured out the logarithm.

This puzzle is hard to solve unless, you heard that limmerick before.

Edited by Prime
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