Let's Compare 3^π and π^3
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- čas přidán 7. 02. 2024
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x^(1/x) is decreasing for x>e. Since pi>3>e, 3^(1/3) > pi^(1/pi). Thus 3^pi > pi^3
Educational as always! If you don't mind, would you be able to make videos covering topics on probability or geometry? That would be awesome, since I struggle in both of those. But Thanks for you videos regardless!
Great suggestion!
The same function (x^(1/x)) can also be used to compare e^pi and pi^e
Really cool vid!
Glad you think so!
You showed that 3^1/3 is greater than π^1/π but I would like to have seen it put back to the original terms to definitively show that 3^π > π^3. I feel that the last step is missing.
I agree. Sorry
Yay, I guessed correctly!
3^(1/3) is the cube root of 3 and is definitely not transcendental.
Lets find minimum of x^x please
Calculus is not maths gift to physics, but it is physics gift to math. It was a physicst who invented calculus, namely Isaac Newton.
Newton, well-known non-mathematician
@@avaraportti1873 Newton, the well-known physicst.
@geraldsmith6225 Physics was the inspiration for developing calculus in the first place.
These problems have been solved many times. Since 3 is closer to e, then 3^π is greater. The interesting case is when one number is less than e and the other is greater than e. Unfortunately, the solution to that problem uses the Lambert W function which requires a calculator to evaluate. I don’t know of a better solution for that case.
I would like three pies.
:D
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@ SyberMath -- 3^(pi) vs. (pi)^3
(pi)*ln(3) vs. 3*ln(pi)
(pi)/3 vs. [ln(pi)]/[ln(3)]
Let 1 + a = (pi)/3, where a < 1, for convenience sake.
Continuing:
1 + a vs. ln[3(1 + a)]/[ln(3)]
1 + a vs. [ln(3) + ln(1 + a)]/[ln(3)]
1 + a vs. [ln(3)]/[ln(3)] + [ln(1 + a)]/[ln(3)]
1 + a vs. 1 + [ln(1 + a)]/[ln(3)]
a vs. [ln(1 + a)]/[ln(3)]
e^x > x + 1 for x > 0 or, equivalently,
x > ln(x + 1) for x > 0.
The numerator ln(1 + a) < a, for a > 0.
Also, the denominator ln(3) > 1.
So, a > [ln(1 + a)]/[ln(3)].
Thus, 3^(pi) > (pi)^3.
Nice
⌈e⌉ = ⌊𝝅⌋ = 3 , but also⌈𝝅^e⌉=⌊e^𝝅⌋ which I find amazingly useless