New answers tagged

1

It seems difficult to disprove, but also extremely improbable. Following Alex R.'s answer, if it were true then we'd have $e = 10^k \pi - A $ for some positive integers $k,B$. And if were true also in other base (say, 7) we'd also have $e = 7^j \pi - B $ for some other integers $j,B$. But that would imply that $\pi$ is rational. Hence, it cannot be true in ...


4

Even if $\pi$ is normal, it doesn't necessarily contain infinitely long patterns. However, I think your question is ultimately unknown: My interpretation of this is: let $\pi(i)$ be the $i$'th digit of $\pi$ $(i=0,1,2,\cdots)$ indexed so that $\pi(0)=3,\pi(1)=1,\pi(2)=4,\ldots$. Similarly, let $e(i)$ by the digits of $e$. You're asking if there exists a $k$ ...


12

No, you cannot. Although $\pi$ has "reasonably concrete" definitions in terms of $+, \times, <$ (e.g. via infinite series), none of them can be made first-order. This follows, e.g., from the fact that: The algebraic reals form a real closed field. The theory of real closed fields is complete, and in fact if $F_1, F_2$ are real closed fields with $F_1\...


1

You can use this method that counts digit for digit in any base: http://bellard.org/pi/pi_n2/pi_n2.html And if you can smuggle in a computer there are programs using this method: http://stackoverflow.com/questions/5905822/function-to-find-the-nth-digit-of-pi


1

Nilakantha's Series haven't been mentioned yet. $$ \pi = 3 + \frac4{2\times3\times4} - \frac4{4\times5\times6} + \frac4{6\times7\times8} - \frac4{8\times9\times10} + \dots $$ Without dots: $$ \pi = 3 + \sum_{k=1}^\infty (-1)^{k+1}\frac1{k \times (2k+1) \times (k+1)} $$ This formula is better than Leibniz's formula, but inefficient compared to Machin's ...


9

An "infinite sum" is not a sum. An infinite sum is the limit of a sequence : $$\sum_{n=0}^\infty=\lim_{N\to\infty} \sum_{n=0}^N.$$ An the limit of a sequence of rational numbers is not necessarily rational.


3

Take any known irrational number, $x$. You can always represent $x$ by the sum of an infinite number of rational numbers. For example, take the decimal representation of $x$: $x = 5.1938527\ldots$ and then each term in the sum could form one of the decimal digits: $x =5 + \frac{1}{10} + \frac{9}{100} + \frac{3}{1000} + \cdots$ Therefore the sum of an ...


1

I have a promising idea but still have to work the details. There is a class of integrals that connect $\pi$ and $e$ that come from: $$ \int_{0}^{+\infty}\frac{\cos(x)}{1+x^2}=\frac{\pi}{2e}\tag{1} $$ Now we may apply integration by parts multiple times, reaching: $$ \frac{\pi}{e} = \int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^k}\,dx \tag{2}$$ with $p(x)$...


9

Here is an answer based upon the arcsine function. We start with the following formula valid for $u\in(0,2)$ \begin{align*} \sum_{n=0}^\infty&\frac{2^{n-1}}{(2n+1)(2n+3)\binom{2n}{n}}u^n\\ &=\frac{1}{u}-\frac{1}{u}\sqrt{\frac{2}{u}-1}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\\ &=\frac{1}{6}+\frac{1}{30}u+\frac{1}{105}u^2+\frac{1}{315}u^3+\...


-1

You don't actually have to establish it. You just note that it is that way. A Turing machine could establish it; don't degrade yourself.


0

The approximations $ 3.14 < \pi < 3.34 $ and $ 2.70 < e < 2.76 $ suffice when doing the computation with two decimals.


3

You can use the fact that $$\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}=2\cos\frac{\pi}{2^{n+1}},$$ where $n$ is the number of radicals in LHS. Thus your sequence is $$2^n\sqrt{2-2\cos\frac{\pi}{2^n}}=2^{n+1}\sin\frac{\pi}{2^{n+1}}\to \pi$$ as $n\to\infty$, since $$\lim_{x\to 0}\frac{\sin\pi x}{x}=\pi.$$


2

Start with \begin{equation} \int \limits_0^x \frac{dt}{1+t^{2}}=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots\qquad + \frac{x^{4n+1}}{4n+1}-R_{n}(x), \quad x \in \mathbb{R}: 0\leq x \leq 1 \end{equation} Where \begin{equation} R_{n}(x) = \int \limits_0^x\frac{t^{4n+2}}{1+t^{2}}dt \end{equation} NOo, since the square of a real number is certainly ...


6

This is a really good question. This issue is very often ignored in online resources, which makes the proofs incomplete. The key here is Abel's theorem. It states that if the function $F(x)$ is defined by a power series $$\sum_{n=1}^\infty a_nx^n$$ on the interval $(-1,1)$ and the series $$\sum_{n=1}^\infty a_n$$ converges to a number $A$, then the limit $$\...


1

Yes, you need an extra step. You need to know this: If $$ \arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots $$ holds for $-1 < x < 1$ and the series on the right side converges at $x=1$, then the equation also holds at $x=1$. Arthur beat me to doing it.


3

Here is a less hand-wavey formulation of what they're saying (as far as I can tell): We have two functions $f, g:(-1, 1) \to \Bbb R$ given by $$ f(x) = \frac1{1+x^2}\\ g(x) = 1-x^2 + x^4 - x^6 + \cdots $$ an they happen to be equal. Note that for any $a \in (-1,1)$, this means that $$ \left(\vphantom{\int}\arctan(a) =\right) \int_0^a f(x)\, dx = \int_0^ag(x)\...


25

The integrand can be broken up as $$I=\int_0^{\infty} \left(\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x} +\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln x}-\frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x}\right)dx.$$ But, by integration by parts, $$\int \frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x} dx= \int \frac{2\ln(1+x)}{x^{1/2} \ln x} d\left(...


21

A preliminary result. $$I_1=\int_{0}^{+\infty}\frac{2\log\left(\frac{1+x}{2}\right)}{\sqrt{x}(1+x)\log(x)}=\color{red}{\pi}.\tag{1}$$ Proof: through the substitution $x=e^t$, $$\begin{eqnarray*}I_1=\int_{-\infty}^{+\infty}\frac{\log\left(\frac{e^t+1}{2}\right)}{\cosh\left(\frac{t}{2}\right)t}\,dt&=&\color{red}{\frac{1}{2}\int_{-\infty}^{+\infty}...


7

My guess is that the denominator is supposed to be $$x^{12} + 6x^{10} + 15x^8 + 20x^6 + 15x^4 + 6x^2 + 1 = (x^2 + 1)^6$$ and the coefficient $20$ was somehow added to the wrong place. In that case, your integral is just $$\int_0^{\infty} \frac{16x^2}{(x^2 + 1)^3} \, \mathrm{d}x,$$ which is easy to solve by substituting $x = \tan(\theta)$: $$\int_0^{\pi/2} \...


0

Use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ and $$2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\left(f(x)+f(a+b-x)\right)dx$$ $$2I=\int_0^{\pi/2}\dfrac{\cos x+\sin x}{2-\sin2x}dx$$ As $\int(\cos x+\sin x)\ dx=\sin x-\cos x,$ let $\sin x-\cos x=u\implies u^2=1-\sin2x$ Can you take it from here?


1

As always, with me, please double check. The sequence is quite obvious. We will transform both to differential equations by the "Method of Coefficients". This is similar to one of Benoit Cloitre's methods but a little more direct. Using the two OGF forms: $$V\left(x\right)={\displaystyle \sum_{k=0}^{\infty}} v_{k}x^{k}, U\left(x\right)={\displaystyle \...


4

Here is a step by step approach. :) $$\begin{align} I &= \int_0^{\pi/2}\frac{\cos{x}}{2-\sin{2x}}dx \\ &= \int_0^{\pi/2}\frac{\cos{x}}{2-2 \sin x \cos x}dx \\ &= \int_0^{\pi/2}\frac{\cos{x}}{1+\cos^2 x -2 \sin x \cos x + \sin^2 x}dx \\ &= \int_0^{\pi/2}\frac{\cos{x}}{1+(\cos x - \sin x)^2}dx \\ &= \frac{1}{2} \left( \int_0^{\pi/2}\frac{\...


2

Hint: 1) $\sin 2x=2\sin x \cos x$ 2) $\sin x =\frac {2t}{1+t^2}$ $\cos x =\frac {1-t^2}{1+t^2}$ $dx=\frac{2dt}{1+t^2}$


12

Hint: Knowing that $\sin2x=2\sin x\cos x$ and $\sin^2x+\cos^2x=1$. The integral can be expressed as \begin{equation} I=\int_0^{\pi/2}\frac{\cos x}{1+(\sin x-\cos x)^2}\ dx \end{equation} then use substitution $x\mapsto\frac{\pi}{2}-x$, we have \begin{equation} I=\int_0^{\pi/2}\frac{\sin x}{1+(\sin x-\cos x)^2}\ dx \end{equation} Add the two $I$'s and ...


3

I write simplify. $$ =\int\frac{d\sin(x-\pi /4)}{ 2 \sin^2(x-\pi/4) +1 } $$ Before it, use $ u=\pi/2 $ to get numerator $\sin x $ and $\cos x$ is same value.


2

Precious Nilakantha's formula! In general from $f(x)$ if you have some value for which $af(x_0)=\pi$ you can developpe $f(x)$ in series and you get a formula $$\pi=a\sum a_nx_0^n$$ This general viewpoint could give some troubles, in particular the value of $x_0$ could be not good, the series cannot be convergent or not useful because convergence very slow (...


0

Take the roots against $\pi e$, to get $e^{1/e}$ v $\pi^{1/\pi}$. These are instances if $x^{1/x}$ for $x=e$ and $x=\pi$. For integers, we see that 9>8 and 2^4=4^2=16 imply that $\sqrt[3]{3} > \sqrt[4]{4} = \sqrt{2}$, from the sixth and eighth roots respectively. Put $y=\sqrt[x]{x}$, whence $x = y^x$. We note that for values greater than 1, there are ...


3

The answer depends on where you want to start. You can start with the definition of $\arctan x$, develop the series expansion for that, and then plug in $x=1$ to set the whole thing equal to $\pi/4$. Or you can calculate it geometrically the way Leibniz did. Absent equating the series to some multiple of $\pi$ you can calculate two series that approach $\...



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