1

How we calculate $2^\pi$? since $\pi$ is irrational how shall I calculate this? and can we write $$(2^\pi)(2^\pi)=(2^\pi)^2$$ and if yes what will be the condition, since $\pi$ is irrational no.

Asaf Karagila
  • 393,674
Mahendra Singh Adhikari
  • 33
  • 1
    $2^\pi$ doesn't means that $$\underbrace{2 \cdot 2 \cdot \dots \cdot 2}_{\pi-\textrm{times}}.$$ – azif00 Aug 27 '20 at 06:12
  • 1
    Will you elaborate this, please ? – Mahendra Singh Adhikari Aug 27 '20 at 06:14
  • 1
    F.e. you may do the following: $\exp(x) \approx 1 + x + x^2/2! + x^3/3! + x^4/4! + O(x^5)$, $\log{2} \approx 1 - 1/2 + 1/3 - 1/4 + 1/5 $ and note $2^\pi = \exp{(\log{2} \cdot \pi)}$ – openspace Aug 27 '20 at 06:16
  • How can we visualise 2^π ? How we can understand it? as we can understand 2^5 means 2 multiplied up to 5 times , yes it can be understand easily because 5 is a natural no. But it is difficult to understand 2^π in same manner , for me .. please explain this – Mahendra Singh Adhikari Aug 27 '20 at 06:20
  • 2
    I disagree with @azifmedrano. $2^{\pi}$ does mean multiplying $\pi$ copies of $2$. This statement would be unambiguously true if $\pi$ were a rational number, say $22/7$. $$2^\pi \approx 2^{22/7} = 2^3\cdot 2^{1/7}$$ This is best described as three and one seventh multiplicative portions of $2$. Since $\pi$ is irrational, we have to massage the definition a little bit and multiply an infinite sequence of ever smaller multiplicative portions of $2$. How many portions? Well, $\pi$ portions! – David Diaz Aug 27 '20 at 06:46
  • If you know $log_{10}(2)$ ($.30103$) then you can easily get a close value. – Déjà vu Aug 27 '20 at 06:46
  • 1
    The "meaning" of $2^{\pi}$ depends on the context, but one way of thinking about it is that the exponential function for positive integers can be extended to all integers and then to rational numbers. When we extend to the real numbers we ask "what functions can we extend, and how?" and from this emerges $2^{\pi}$. In fact the exponential function (with $e$ as base) turns out to be very useful (as a fixed point of the differentiation operation). But the extension from rationals to reals is itself non-trivial. – Mark Bennet Aug 27 '20 at 07:22
  • 4
    Your question was already asked here: Exponent of a number is a square root? – user21820 Aug 10 '22 at 10:27
  • @MarkBennet: Exactly, see the linked post. I don't like the fact that most answers here just say things without any proper motivation or justification. – user21820 Aug 10 '22 at 10:29

4 Answers4

3

$$A=2^\pi=2^3\, 2^{\pi-3}=8 \, 2^{\pi-3}=8\, \exp\big[(\pi-3)\log(2) \big]$$ Now, for small $x$, a good approximation is $$e^x=\frac{2+x}{2-x}$$ and $(\pi-3)\log(2)\sim 0.098$ $$A\sim 8 \,\frac{2+0.098}{2-0.098}=\frac{8392}{951}\sim 8.8244$$ while the exact result is $8.8250$.

Claude Leibovici
  • 260,315
2

If $n$ is a positive integer:

  • $2^n$ means the product of $n$ copies of $2$,
  • $2^{-n}$ means $(2^n)^{-1} = (2^{-1})^n$ (why this are the same number?),
  • $2^{\frac 1n}$ means $\sqrt[n]{2}$, and
  • $2^{-\frac 1n}$ means $(2^{\frac 1n})^{-1} = (2^{-1})^{\frac 1n}$ (again, why?).

Also, $2^0$ is $1$. Now, if $p$ and $q$ are two integers with $q \neq 0$, $2^{\frac pq}$ means $(2^{\frac 1q})^p = (2^p)^{\frac 1q}$ (?). With all of this, we know the value of $2^r$ for any rational number $r$, right? Finally, if $x$ is a real number, choose a sequence of rational numbers $r_1,r_2,\dots$ such that $\lim_n r_n = x$, and then $2^x$ means $\lim_{n} 2^{r_n}$.

azif00
  • 20,792
  • √2 means side of a square whose area is 2 . And cuberoot of 2 means side of a cube whose volume is 2 in the same manner please kindly explain the meaning of 2^π please . – Mahendra Singh Adhikari Aug 27 '20 at 06:39
  • Yes, all of that you mention is covered in $2^{\frac 1n} = \sqrt[n]{2}$. – azif00 Aug 27 '20 at 06:46
  • @MahendraSinghAdhikari the point of this answer is exactly to argue that the definition of $2^{\pi}$ isn't expressed in terms of sides of plain geometrical figures. – gented Aug 27 '20 at 14:41
1
  1. To answer your motivating question ("How can we visualise $2^\pi$? How we can understand it?") in the 4th comment under the main question:

$\\ a^x$ is defined as $e^{x\ln a}$ for all positive $a$ and real $x$.

$e^y$ is defined as $\lim_{n\to\infty}\left(1+\frac y{n}\right)^n$ for all real $y$.

Therefore, by definition, $$2^\pi=\lim_{n\to\infty}\left(1+\frac {\pi\ln 2}{n}\right)^n.$$ Equivalently (from the power series expansion of $e^{\pi\ln 2}$), $$2^\pi=\sum_{n=0}^{\infty}\frac{(\pi\ln 2)^n}{n!}.$$

  1. Plugging in the power series expansions of $(4\arctan1=)\pi$ and $\ln 2$ gives \begin{equation} \begin{split} 2^\pi=\sum_{n=0}^{\infty}\frac 1{n!}&\left[4\left(\sum_{n=0}^{\infty}(-1)^n\frac1{2n+1}\right)\left(\sum_{n=1}^{\infty}(-1)^{n+1}\frac1{n}\right)\right]^n \\ =1+\frac11&\left[4\left(\frac 11-\frac 13+\frac 15-\ldots\right)\left(\frac 11-\frac12+\frac13-\dots\right)\right] \\ +\frac 1{2!}&\left[4\left(\frac 11-\frac 13+\frac 15-\ldots\right)\left(\frac 11-\frac12+\frac13-\dots\right)\right]^2 \\ +\frac 1{3!}&\left[4\left(\frac 11-\frac 13+\frac 15-\ldots\right)\left(\frac 11-\frac12+\frac13-\dots\right)\right]^3 \\ +\ldots. \end{split} \end{equation} This theoretically lets us estimate $2^\pi$; however it converges too slowly to be of practical use.
ryang
  • 38,879
  • 14
  • 81
  • 179
1

Here will make use of calculus approximation using differentials

$$ y= 2^x \\ \ln\ y = x\ln\ 2 \\ \frac{1\ dy}{y\ dx}\ = \ln\ 2 \\ dy = y\ln\ 2\ dx \\ dy = 2^x \ln\ 2\ dx \\ $$ now use a differential approach $$ \Delta y = 2^x\ln 2\ \Delta x $$ Now, using $x=3, \ \Delta x = 0.14159265359$ we see that $\Delta y=78159144782$, that added to the original value in $2^3=8$ is $8.781581$ which is close to the value, although not as close as other methods shown here.

Asaf Karagila
  • 393,674
Favio Villaseñor
  • 75