Math Is Fun Forum

Hi ganesh,
The question M#169 originally stated that the base area of the cone was 12 (I assume square centimetres), not the radius.

Pshaw! Just because the bottom of the cylinder rests on the ground, it does not count as part of the surface area?

This is going to sound really wacky and unschooled, but if there are two limits that predict 0^x = 0 if x = 0 and y^0 = 1 if y = 0, instead of throwing one's hands up, could they average the two?

Hello!
I think diagrams would be helpful for most of these xD They turn into interpretation problems just as much as geometry problems.
I assume that we want the curved surface area of the cone, plus the open surface area of the cylinder (i.e. without the top), where the radius is in common.

Hello, I cannot exactly give an in-depth explanation, but basically I used a lot of doubling and adding 1 in Excel to generate all of the sets up to S20, summed each of them, and posted the sums so somebody else could have a look at them before I came back to the problem of the general rule for the sums.

Then bobbym waved his experimental wand on the numbers, and we get the long rule in post #17, which, I have confirmed, does in fact work for the sums that I have. It is obviously not meant for an assignment, considering that every other question was quite straightforward but the rule for the sum of Sn is that thing.
If you were after a proof for that rule in #17, it is my understanding that it is unproven, but confirmed to an uncertain but high probability.

LOL to both of those xD

Sheldon: Why are you crying?
Penny: Because I'm stupid!
Sheldon: Well that's no reason to cry; one cries because one is sad. For example, I cry because others are stupid and it makes me sad.

Leonard: I’m having dinner with Priya at Raj’s. I think Howard’s going to be there. You want to join us?
Sheldon: But tonight’s Thursday. On Thursdays, everybody comes over here and has pizza.
Leonard: Can’t we make a one-time exception for tonight?
Sheldon: We could. We could also stop using the letter M, but I think that idea is isguided and oronic.

Sheldon: Okay, Penny. That's your last chance to prove to me you're not stupid.
Penny: Okay, Sheldon. I'll prove to you I'm not stupid.
Sheldon: Mary's father has 5 children - 1. Nana 2. Nene 3. Nini 4. Nono. What is the name of the fifth child?
Penny: Nunu!
Sheldon: Mary!
Penny: O:\

Stuart: You guys still on for bowling tonight?
Sheldon: Oh, yes. In fact, I've prepared some trash-talk for the occasion. Ahem... You bowl like your momma!
[short pause]
Sheldon: Unless she bowls well, in which case you bowl nothing like her.
Stuart: Oh. Ouch.

Leonard: For God's sake, Sheldon, do I have to hold up a sarcasm sign every time I open my mouth?
Sheldon: You have a sarcasm sign?

I like the idea of zeroing in on an answer very much. Why can you not specify what the probability of it being true is?

And for a bit of fun, an answer with this formula can be given to the hypothetical question in post #6: What is the sum of S72?
According to the formula, it is: 26,741,856,567,998,641,197,432,083,722,543 or about 26.7 nonillion.

The largest term is 4,722,366,482,869,645,213,696 or about 4.7 sextillion.
The smallest term is 137,438,953,471 or about 137.4 billion.

It has 806,515,533,049,393 or about 806.5 trillion terms.
498,454,011,879,264 or about 498.45 trillion of them are even.
308,061,521,170,129 or about 308 trillion of them are odd.

Furthermore, the average term is about 33 quadrillion.

Hm... Well I personally prefer full disclosure, but if required I will foolishly follow my own methods while you move on to greater things like predicting the nth decimal point of pi to the pi or something.

Oh I see, I must have misapplied something. Interestingly, however, it seems that using 3.14 ~ pi gets you an answer of about 7.211102551cm, while using the exact value gives about 7.744657513cm. The discrepancy is either because of the compounding squaring and square-rooting going on, or because I am again wrong xD

Hi bob;

I have confirmed your formulas for sets up to S20, which is ample confirmation.

I suspect that I will give this problem another good go quite soon, but not just yet. In the meantime, though, I have brute force calculated all of the sums up to and including S20 for scrutiny, which was a big task in itself.

S1: 2
S2: 7
S3: 19
S4: 54
S5: 149
S6: 411
S7: 1128
S8: 3091
S9: 8459
S10: 23,134
S11: 63,241
S12: 172,839
S13: 472,304
S14: 1,290,519
S15: 3,526,023
S16: 9,633,694
S17: 26,320,421
S18: 71,909,827
S19: 196,463,080
S20: 536,749,995

Notice the ratios. They appear to converge to a number roughly 2.732076. Can this number, or a number slightly below it, be connected to Fibonacci? Or is it something else altogether? (The best I've got is that phi^2 ~ 2.618033989)

Hello;
The exact answers for slant height, height and radius are a bit complicated, even with the 22/7 ~ pi approximation. However, to ten significant figures, the answers I get are:

Hi;