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Aaron Meyerowitz
Aaron Meyerowitz
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We can obtain such an infinity of sides of squares by continuously increasing the length of the side of the inscribed square to the length of the circumscribed square of a circle with diameter equal to one. The first new square is obtained by taking the square root of the arithmetic mean of the sides of the inscribed and circumscribed squares so we have $\frac{1+\frac{sqrt{2}}{2}}{2}=\frac{\sqrt{2+\sqrt2}}{2}$$\frac{1+\frac{\sqrt{2}}{2}}{2}=\frac{\sqrt{2+\sqrt2}}{2}$. The side of the second new square is obtained by taking the square root of the arithmetic mean of $1$ and $\frac{\sqrt{2+\sqrt2}}{2}$. We can continue this process indefinitely. Since an infinite product of sides of the circumscribed squares converges, so does the infinite product of the squares we obtain by the above process. This product converges to a transcendental number.

Another infinite product of sides of squares is obtained as follows. Let's have a straight line $AC$ and $AB=BC=1$. From point $A$ to $B$ we can construct an infinity of sides of squares by the continuous bisection of the length $AB$. If $n=1$ we have $AB_1=\frac{1}{2}$ if $n=2$ we have $B_1B_2=\frac{1}{4}$ and so on indefinitely. The sides of squares which we want are obtained as follows. $AC_n=AB+\frac{1}{2^n}$. If $n=1$ we have $AC_1=\frac{3}{2}$ if $n=2$ we have $AC_2=\frac{5}{4}$ and so on. Each product of the side of these squares is obtained by multiplying the side $2^{n-1}$ times. When $n=2$ we have the product $\frac{5}{4}\star\frac{5}{4}=\frac{25}{16}$ if $n=3$ we have the product $\frac{9}{8}\star \frac{9}{8}\star\frac{9}{8}\star \frac{9}{8}=\frac{6561}{4096}$ and so on. Since the bisections go on indefinitely so does the product of the side of the square $AC_n$. So the number to which such a product converges cannot be a constant quantity of an algebraic equation because algebraic equations require a finite number as a constant quantity. But with the above process we cannot have a final side or a final product either. My first question is: Is there any other transcendental which can be obtained as an infinite product of sides of squares? Second question: If we take the square root of the arithmetic mean of $1$ and $\frac{1}{2}$, equal to $\frac{\sqrt3}{2}$, and then the square root of the arithmetic mean of $1$ and $\frac{\sqrt3}{2}$ and so on indefinitely the product of these numbers converges but I could not find a closed form of this number. Does anyone know to express this product of numbers in closed form?

We can obtain such an infinity of sides of squares by continuously increasing the length of the side of the inscribed square to the length of the circumscribed square of a circle with diameter equal to one. The first new square is obtained by taking the square root of the arithmetic mean of the sides of the inscribed and circumscribed squares so we have $\frac{1+\frac{sqrt{2}}{2}}{2}=\frac{\sqrt{2+\sqrt2}}{2}$. The side of the second new square is obtained by taking the square root of the arithmetic mean of $1$ and $\frac{\sqrt{2+\sqrt2}}{2}$. We can continue this process indefinitely. Since an infinite product of sides of the circumscribed squares converges, so does the infinite product of the squares we obtain by the above process. This product converges to a transcendental number.

Another infinite product of sides of squares is obtained as follows. Let's have a straight line $AC$ and $AB=BC=1$. From point $A$ to $B$ we can construct an infinity of sides of squares by the continuous bisection of the length $AB$. If $n=1$ we have $AB_1=\frac{1}{2}$ if $n=2$ we have $B_1B_2=\frac{1}{4}$ and so on indefinitely. The sides of squares which we want are obtained as follows. $AC_n=AB+\frac{1}{2^n}$. If $n=1$ we have $AC_1=\frac{3}{2}$ if $n=2$ we have $AC_2=\frac{5}{4}$ and so on. Each product of the side of these squares is obtained by multiplying the side $2^{n-1}$ times. When $n=2$ we have the product $\frac{5}{4}\star\frac{5}{4}=\frac{25}{16}$ if $n=3$ we have the product $\frac{9}{8}\star \frac{9}{8}\star\frac{9}{8}\star \frac{9}{8}=\frac{6561}{4096}$ and so on. Since the bisections go on indefinitely so does the product of the side of the square $AC_n$. So the number to which such a product converges cannot be a constant quantity of an algebraic equation because algebraic equations require a finite number as a constant quantity. But with the above process we cannot have a final side or a final product either. My first question is: Is there any other transcendental which can be obtained as an infinite product of sides of squares? Second question: If we take the square root of the arithmetic mean of $1$ and $\frac{1}{2}$, equal to $\frac{\sqrt3}{2}$, and then the square root of the arithmetic mean of $1$ and $\frac{\sqrt3}{2}$ and so on indefinitely the product of these numbers converges but I could not find a closed form of this number. Does anyone know to express this product of numbers in closed form?

We can obtain such an infinity of sides of squares by continuously increasing the length of the side of the inscribed square to the length of the circumscribed square of a circle with diameter equal to one. The first new square is obtained by taking the square root of the arithmetic mean of the sides of the inscribed and circumscribed squares so we have $\frac{1+\frac{\sqrt{2}}{2}}{2}=\frac{\sqrt{2+\sqrt2}}{2}$. The side of the second new square is obtained by taking the square root of the arithmetic mean of $1$ and $\frac{\sqrt{2+\sqrt2}}{2}$. We can continue this process indefinitely. Since an infinite product of sides of the circumscribed squares converges, so does the infinite product of the squares we obtain by the above process. This product converges to a transcendental number.

Another infinite product of sides of squares is obtained as follows. Let's have a straight line $AC$ and $AB=BC=1$. From point $A$ to $B$ we can construct an infinity of sides of squares by the continuous bisection of the length $AB$. If $n=1$ we have $AB_1=\frac{1}{2}$ if $n=2$ we have $B_1B_2=\frac{1}{4}$ and so on indefinitely. The sides of squares which we want are obtained as follows. $AC_n=AB+\frac{1}{2^n}$. If $n=1$ we have $AC_1=\frac{3}{2}$ if $n=2$ we have $AC_2=\frac{5}{4}$ and so on. Each product of the side of these squares is obtained by multiplying the side $2^{n-1}$ times. When $n=2$ we have the product $\frac{5}{4}\star\frac{5}{4}=\frac{25}{16}$ if $n=3$ we have the product $\frac{9}{8}\star \frac{9}{8}\star\frac{9}{8}\star \frac{9}{8}=\frac{6561}{4096}$ and so on. Since the bisections go on indefinitely so does the product of the side of the square $AC_n$. So the number to which such a product converges cannot be a constant quantity of an algebraic equation because algebraic equations require a finite number as a constant quantity. But with the above process we cannot have a final side or a final product either. My first question is: Is there any other transcendental which can be obtained as an infinite product of sides of squares? Second question: If we take the square root of the arithmetic mean of $1$ and $\frac{1}{2}$, equal to $\frac{\sqrt3}{2}$, and then the square root of the arithmetic mean of $1$ and $\frac{\sqrt3}{2}$ and so on indefinitely the product of these numbers converges but I could not find a closed form of this number. Does anyone know to express this product of numbers in closed form?

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Vassilis Parassidis
Vassilis Parassidis
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Transcendental numbers as infinite products of sides of squares

We can obtain such an infinity of sides of squares by continuously increasing the length of the side of the inscribed square to the length of the circumscribed square of a circle with diameter equal to one. The first new square is obtained by taking the square root of the arithmetic mean of the sides of the inscribed and circumscribed squares so we have $\frac{1+\frac{sqrt{2}}{2}}{2}=\frac{\sqrt{2+\sqrt2}}{2}$. The side of the second new square is obtained by taking the square root of the arithmetic mean of $1$ and $\frac{\sqrt{2+\sqrt2}}{2}$. We can continue this process indefinitely. Since an infinite product of sides of the circumscribed squares converges, so does the infinite product of the squares we obtain by the above process. This product converges to a transcendental number.

Another infinite product of sides of squares is obtained as follows. Let's have a straight line $AC$ and $AB=BC=1$. From point $A$ to $B$ we can construct an infinity of sides of squares by the continuous bisection of the length $AB$. If $n=1$ we have $AB_1=\frac{1}{2}$ if $n=2$ we have $B_1B_2=\frac{1}{4}$ and so on indefinitely. The sides of squares which we want are obtained as follows. $AC_n=AB+\frac{1}{2^n}$. If $n=1$ we have $AC_1=\frac{3}{2}$ if $n=2$ we have $AC_2=\frac{5}{4}$ and so on. Each product of the side of these squares is obtained by multiplying the side $2^{n-1}$ times. When $n=2$ we have the product $\frac{5}{4}\star\frac{5}{4}=\frac{25}{16}$ if $n=3$ we have the product $\frac{9}{8}\star \frac{9}{8}\star\frac{9}{8}\star \frac{9}{8}=\frac{6561}{4096}$ and so on. Since the bisections go on indefinitely so does the product of the side of the square $AC_n$. So the number to which such a product converges cannot be a constant quantity of an algebraic equation because algebraic equations require a finite number as a constant quantity. But with the above process we cannot have a final side or a final product either. My first question is: Is there any other transcendental which can be obtained as an infinite product of sides of squares? Second question: If we take the square root of the arithmetic mean of $1$ and $\frac{1}{2}$, equal to $\frac{\sqrt3}{2}$, and then the square root of the arithmetic mean of $1$ and $\frac{\sqrt3}{2}$ and so on indefinitely the product of these numbers converges but I could not find a closed form of this number. Does anyone know to express this product of numbers in closed form?