Pies all over the floor!

Hello everybody! This is my very first note. As the title suggests, today we are going to talk about something related to pi. Nah, not the chicken pie or apple pie or any other pies that you enjoy eating but the pi defined as the ratio of circumference to the diameter of a circle!

Well, to get things started, find \(\lfloor \pi\rfloor+\lfloor \pi^{2}\rfloor+\lfloor \pi^{3}\rfloor+...+\lfloor \pi^{10}\rfloor\) without using a calculator. Actually, this is a question which I propose yet fail to solve until now.

More generally, find \(\lfloor \pi\rfloor+\lfloor \pi^{2}\rfloor+\lfloor \pi^{3}\rfloor+...+\lfloor \pi^{n}\rfloor\) in terms of \(n\).

I hope someone can offer help or suggestions. If you find this interesting just like I do, help me like or re share this note to attract the attention of pros, thank you. I'll sign off here for now.

Cheers!

Note by Donglin Loo
3 months, 2 weeks ago

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1 vote

  Easy Math Editor

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Comments

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Perhaps utilize \(\dfrac{\pi}{2}=\dfrac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot8\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdot9\cdots}\)?

Kenneth Tan - 3 months, 2 weeks ago

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Wow. Is there a proof for this?

Donglin Loo - 3 months, 1 week ago

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Wallis Product

X X - 3 months, 1 week ago

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@X X Thanks @X X

Donglin Loo - 3 months, 1 week ago

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Take a look at this OEIS ,though it may not be helpful.

X X - 3 months, 2 weeks ago

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Oh I see. Thanks for your idea

Donglin Loo - 3 months, 1 week ago

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Interesting question, may use this identity \[1+\dfrac14+\dfrac19+\dfrac1{16}+\cdots=\dfrac{\pi^2}6\] to do that but I'm not sure...

Kelvin Hong - 3 months, 2 weeks ago

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Oh wow. I did not think of this. Thanks for your suggestion.

Donglin Loo - 3 months, 1 week ago

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\(\pi = 3.14 \implies \lfloor \pi\rfloor= 3\)

So, we have to find \(\lfloor \pi\rfloor+\lfloor \pi^{2}\rfloor+\lfloor \pi^{3}\rfloor+...+\lfloor \pi^{10}\rfloor\)

\(\implies 3 + 3^2 + 3^3 + 3^4 + 3^5 .... + 3^{10}\)

The above series is in G.P with \(a = 3, r = 3, n = 10\)

Sum of terms in G.P = \(\dfrac{3(3^{10} - 1)}{3 - 1}\)

\(\implies 1.5(3^{10} - 1)\)

Coming to the next part of the question sum of infinite terms in an increasing G.P is infinite.

However, if \(n\) is finite the formula would be :

\(\dfrac{3(3^n -1)}{3 - 1}\)

Ram Mohith - 3 months, 2 weeks ago

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@Ram Mohith. You can't assume \(\pi=3\) just like that. For instance, \(\pi^3=31.000627668\) so \(\lfloor \pi^3\rfloor=31\) instead of \(3^3=27\). You can verify this with a calculator. The difficult part is that \(\pi=3.14159265\) and with exponentiation things can get very random.

Donglin Loo - 3 months, 2 weeks ago

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