Elementary Approach to Modular Equations: Ramanujan's Theory 6

The Fundamental Formulas

In this post we will continue our journey of modular equations and derive a host of these mostly by using Lambert series for various theta functions. The following formula (see equation $(14)$ of this post) will be of great help here: $$\phi^{2}(-ab)\,\frac{f(a, b)}{f(-a, -b)} = 1 + 2\sum_{n = 1}^{\infty}\frac{a^{n} + b^{n}}{1 + a^{n}b^{n}}\tag{1}$$
It is best to establish some properties of the function $f(a, b)$ here. Let $p = ab$ and then we can write $$\begin{align}f(a, b) &= \sum_{n = -\infty}^{\infty}a^{n(n + 1)/2}b^{n(n - 1)/2}\notag\\ &= \sum_{n = -\infty}^{\infty}p^{n(n + 1)/2}b^{-n}\notag\\ &= \sum_{n = -\infty}^{\infty}p^{n(n - 1)/2}a^{n}\notag\end{align}$$ and hence we can write $$\sum_{n = -\infty}^{\infty}x^{n(n + 1)/2}y^{n} = f(xy, 1/y)$$ $$\sum_{n = -\infty}^{\infty}x^{n(n - 1)/2}y^{n} = f(y, x/y)$$ We now establish the following set of identities (here $p = ab$ and $f(q) = (-q;-q)_{\infty}$):

• $\displaystyle \prod_{k = 1}^{n}f(ap^{k - 1}, bp^{n - k}) = \frac{f(a, b)f^{n}(-p^{n})}{f(-p)}$
• $\displaystyle f(a, ab^{2})f(b, a^{2}b) = f(a, b)\psi(ab)$
• $\displaystyle f(a, b) + f(-a, -b) = 2f(a^{3}b, ab^{3})$
• $\displaystyle f(a, b) - f(-a, -b) = 2af(b/a, (a/b)a^{4}b^{4})$
• $\displaystyle f(a, b)f(-a, -b) = f(-a^{2}, -b^{2})\phi(-ab)$
• $\displaystyle f^{2}(a, b) - f^{2}(-a, -b) = 4af(b/a, (a/b)a^{2}b^{2})\psi(a^{2}b^{2})$

In proving the above identities heavy use will be made of the results established in a previous post. Since this will be done implicitly the reader must take note of each step and refer to the results from the previous post if necessary. The first one can be derived as follows: $$\begin{align}\frac{f(a, b)f^{n}(-p^{n})}{f(-p)} &= (-a; p)_{\infty}(-b; p)_{\infty}(p^{n}; p^{n})_{\infty}^{n}\notag\\ &= \left\{\prod_{k = 1}^{n}(-ap^{k - 1}; p^{n})_{\infty}\right\}\left\{\prod_{k = 1}^{n}(-bp^{k - 1}; p^{n})_{\infty}\right\}(p^{n};p^{n})_{\infty}^{n}\notag\\ &= \prod_{k = 1}^{n}(-ap^{k - 1}; p^{n})_{\infty}(-bp^{n - k}; p^{n})_{\infty}(p^{n}; p^{n})_{\infty}\notag\\ &= \prod_{k = 1}^{n}f(ap^{k - 1}, bp^{n - k})\notag\end{align}$$ Putting $n = 2$ in the above identity we get $$\begin{align}f(a, ab^{2})f(b, a^{2}b) &= \frac{f(a, b)f^{2}(-p^{2})}{f(-p)}\notag\\ &= \frac{f(a, b)f^{3}(-p^{2})}{f(-p)f(-p^{2})}\notag\\ &= \frac{f(a, b)\phi(-p)\psi^{2}(p)}{f(-p)f(-p^{2})}\notag\\ &= f(a, b)\psi(p) = f(a, b)\psi(ab)\notag\end{align}$$ For the third one we have $$\begin{align} f(a, b) + f(-a, -b) &= \sum_{n = -\infty}^{\infty}p^{n(n - 1)/2}(a^{n} + (-a)^{n})\notag\\ &= 2\sum_{n = -\infty}^{\infty}p^{n(2n - 1)}a^{2n}\notag\\ &= 2\sum_{n = -\infty}^{\infty}p^{2n^{2} - n}a^{2n}\notag\\ &= 2\sum_{n = -\infty}^{\infty}p^{(4n^{2} - 4n)/2}p^{n}a^{2n}\notag\\ &= 2\sum_{n = -\infty}^{\infty}(p^{4})^{n(n - 1)/2}(a^{2}p)^{n}\notag\\ &= 2f(a^{2}p, p^{4}/(a^{2}p)) = 2f(a^{3}b, ab^{3})\notag\end{align}$$ Similarly the fourth one can be established. Next we have $$\begin{align}f(a, b)f(-a, -b) &= (-a;p)_{\infty}(-b;p)_{\infty}(p;p)_{\infty}(a;p)_{\infty}(b;p)_{\infty}(p;p)_{\infty}\notag\\ &= (a^{2};p^{2})_{\infty}(b^{2};p^{2})_{\infty}(p;p)_{\infty}^{2}\notag\\ &= f(-a^{2}, -b^{2})\cdot\frac{(p;p)_{\infty}^{2}}{(p^{2};p^{2})_{\infty}}\notag\\ &= f(-a^{2}, -b^{2})\phi(-p) = f(-a^{2}, -b^{2})\phi(-ab)\notag\end{align}$$ Multiplying the 3rd and 4th identities above we get $$\begin{align}f^{2}(a, b) - f^{2}(-a, -b) &= 4af(a^{3}b, ab^{3})f(b/a, (a/b)a^{4}b^{4})\notag\\ &= 4af(x, xy^{2})f(y, x^{2}y) = 4af(x, y)\psi(xy)\notag\\ &= 4af(b/a, (a/b)a^{2}b^{2})\psi(a^{2}b^{2})\notag\end{align}$$ where $x = b/a, y = a^{3}b$. Thus all the above identities are established.

Next we have two conditional identities. If $ab = cd$ then

• $\displaystyle f(a, b)f(c, d) + f(-a, -b)f(-c, -d) = 2f(ac, bd)f(ad, bc)$
• $\displaystyle f(a, b)f(c, d) - f(-a, -b)f(-c, -d) = 2af\left(\frac{b}{c}, \frac{c}{b}abcd\right)f\left(\frac{b}{d}, \frac{d}{b}abcd\right)$

Clearly if $p = ab = cd$, then $$\begin{align} f(a, b)f(c, d) &= \sum_{m = -\infty}^{\infty}p^{m(m - 1)/2}a^{m}\sum_{n = -\infty}^{\infty}p^{n(n - 1)/2}c^{n}\notag\\ &= \sum_{m, n = -\infty}^{\infty}p^{(m^{2} + n^{2})/2 - (m + n)/2}a^{m}c^{n}\notag\end{align}$$ and hence we have $$f(a, b)f(c, d) + f(-a, -b)f(-c, -d) = 2\sum_{m, n = -\infty, m + n = 2k}^{\infty}p^{(m^{2} + n^{2})/2 - (m + n)/2}a^{m}c^{n}$$ Setting $m - n = 2j$ we get $$\begin{align}f(a, b)f(c, d) + f(-a, -b)f(-c, -d) &= 2\sum_{j, k = -\infty}^{\infty}p^{j^{2} + k^{2} - k}a^{j + k}c^{k - j}\notag\\ &= 2\sum_{j, k = -\infty}^{\infty}p^{j^{2} + k^{2} - k - j}a^{j + k}c^{k - j}p^{j}\notag\\ &= 2\sum_{j, k = -\infty}^{\infty}p^{j^{2} + k^{2} - k - j}a^{j + k}c^{k - j}(cd)^{j}\notag\\ &= 2\sum_{j, k = -\infty}^{\infty}p^{j^{2} + k^{2} - k - j}(ac)^{j}(ad)^{k}\notag\\ &= 2\sum_{j, k = -\infty}^{\infty}(p^{2})^{(j^{2} + k^{2})/2 - (k + j)/2}(ac)^{j}(ad)^{k}\notag\\ &= 2 f(ac, bd)f(ad, bc)\notag\end{align}$$ Similarly the second conditional identity can be proved. We will use these identities to establish analogues of the fundamental formula $(1)$. Replacing $a$ by $(-a)$ in $(1)$ we get $$\phi^{2}(ab)\,\frac{f(-a, b)}{f(a, -b)} = 1 + 2\sum_{n = 1}^{\infty}\frac{(-a)^{n} + b^{n}}{1 + (-ab)^{n}}$$ Interchanging $a$ and $b$ we get $$\phi^{2}(ab)\,\frac{f(a, -b)}{f(-a, b)} = 1 + 2\sum_{n = 1}^{\infty}\frac{a^{n} + (-b)^{n}}{1 + (-ab)^{n}}$$ Subtracting the above two relations we get $$\begin{align}\phi^{2}(ab)\left\{\frac{f(a, -b)}{f(-a, b)} - \frac{f(-a, b)}{f(a, -b)}\right\} &= 4\sum_{n = 1}^{\infty}\frac{a^{2n - 1} - b^{2n - 1}}{1 - (ab)^{2n - 1}}\notag\\ \Rightarrow \phi^{2}(ab)\cdot\frac{f^{2}(a, -b) - f^{2}(-a, b)}{f(a, -b)f(-a, b)} &= 4\sum_{n = 1}^{\infty}\frac{a^{2n - 1} - b^{2n - 1}}{1 - (ab)^{2n - 1}}\notag\\ \Rightarrow \phi^{2}(ab)\cdot\frac{4af(-b/a, -a^{3}b)\psi(a^{2}b^{2})}{f(-a^{2}, -b^{2})\phi(ab)} &= 4\sum_{n = 1}^{\infty}\frac{a^{2n - 1} - b^{2n - 1}}{1 - (ab)^{2n - 1}}\notag\\ \Rightarrow a\frac{f(-b/a, -a^{3}b)}{f(-a^{2}, -b^{2})}\phi(ab)\psi(a^{2}b^{2}) &= \sum_{n = 1}^{\infty}\frac{a^{2n - 1} - b^{2n - 1}}{1 - (ab)^{2n - 1}}\tag{2}\end{align}$$ Expanding the sum on right as a double sum and then summing by columns we get $$a\frac{f(-b/a, -a^{3}b)}{f(-a^{2}, -b^{2})}\phi(ab)\psi(a^{2}b^{2}) = \sum_{n = 1}^{\infty}\left(\frac{a^{n}b^{n - 1}}{1 - a^{2n}b^{2n - 2}} - \frac{a^{n - 1}b^{n}}{1 - a^{2n - 2}b^{2n}}\right)\tag{3}$$

Relation Between Theta Functions of $q$ and $q^{3}$

Using the above identities we can establish further relations between theta functions of $q$ and $q^{3}$. Putting $a = q, b = -q^{2}$ in $(1)$ we get $$\begin{align}\phi^{2}(q^{3})\frac{f(q, -q^{2})}{f(-q, q^{2})} &= 1 + 2\sum_{n = 1}\frac{q^{n} + (-q^{2})^{n}}{1 + (-q)^{3n}}\notag\\ \Rightarrow \phi^{2}(q^{3})\frac{f^{2}(q, -q^{2})}{f(q, -q^{2})f(-q, q^{2})} &= 1 + 2\sum_{n = 1}\frac{q^{n} + (-q^{2})^{n}}{1 + (-q)^{3n}}\notag\\ \Rightarrow \phi^{2}(q^{3})\frac{f^{2}(q)}{f(-q^{2}, -q^{4})\phi(q^{3})} &= 1 + 2\sum_{n = 1}\frac{q^{n} + (-q^{2})^{n}}{1 + (-q)^{3n}}\notag\\ \Rightarrow \frac{f^{2}(q)}{f(-q^{2})}\phi(q^{3}) &= 1 + 2\sum_{n = 1}\frac{q^{n} + (-q^{2})^{n}}{1 + (-q)^{3n}}\notag\\ \Rightarrow \phi(q)\phi(q^{3}) &= 1 + 2\sum_{n = 1}\frac{q^{n} + (-q^{2})^{n}}{1 + (-q)^{3n}}\notag\end{align}$$ If we sum the series on right by columns (this requires some patience to verify) then we get $$\phi(q)\phi(q^{3}) = 1 + 2\left(\frac{q}{1 - q} - \frac{q^{2}}{1 + q^{2}} + \frac{q^{4}}{1 + q^{4}} - \frac{q^{5}}{1 - q^{5}} + \cdots\right)$$ i.e. $$\phi(q)\phi(q^{3}) = 1 + 2\sum_{n = 0}^{\infty}\left(\frac{q^{6n + 1}}{1 - q^{6n + 1}} - \frac{q^{6n + 2}}{1 + q^{6n + 2}} + \frac{q^{6n + 4}}{1 + q^{6n + 4}} - \frac{q^{6n + 5}}{1 - q^{6n + 5}}\right)\tag{4}$$ Again putting $a = q, b = q^{5}$ in relation $(3)$ we get $$\begin{align} \sum_{n = 0}^{\infty}\left(\frac{q^{6n + 1}}{1 - q^{12n + 2}} - \frac{q^{6n + 5}}{1 - q^{12n + 10}}\right) &= q\frac{f(-q^{4}, -q^{8})}{f(-q^{2}, -q^{10})}\phi(q^{6})\psi(q^{12})\notag\\ &= q\frac{f(-q^{4})(q^{6};q^{12})_{\infty}\psi^{2}(q^{6})}{(q^{2};q^{12})_{\infty}(q^{6};q^{12})_{\infty}(q^{10};q^{12})_{\infty}(q^{12};q^{12})_{\infty}}\notag\\ &= q\frac{f(-q^{4})(q^{6};q^{12})_{\infty}}{(q^{2};q^{4})_{\infty}(q^{12};q^{12})_{\infty}}\psi^{2}(q^{6})\notag\\ &= q\frac{(q^{4};q^{4})_{\infty}(q^{6};q^{12})_{\infty}}{(q^{2};q^{4})_{\infty}(q^{12};q^{12})_{\infty}}\psi^{2}(q^{6})\notag\\ &= q\frac{\psi(q^{2})}{\psi(q^{6})}\psi^{2}(q^{6})\notag\\ &= q\psi(q^{2})\psi(q^{6})\notag\end{align}$$ Thus $$q\psi(q^{2})\psi(q^{6}) = \sum_{n = 0}^{\infty}\left(\frac{q^{6n + 1}}{1 - q^{12n + 2}} - \frac{q^{6n + 5}}{1 - q^{12n + 10}}\right)\tag{5}$$ If we replace $q$ by $(-q)$ in $(4)$ and subtract the resulting equation from $(4)$ we immediately see that $$4q\psi(q^{2})\psi(q^{6}) = \phi(q)\phi(q^{3}) - \phi(-q)\phi(-q^{3})\tag{6}$$

Modular Equation of Degree 3

We can now transcribe this into a modular equation by using transcription formulas from previous post and then we obtain $$4e^{-y}\cdot\frac{1}{2}\sqrt{z}(xe^{y})^{1/4}\frac{1}{2}\sqrt{z_{1}}(x_{1}e^{y_{1}})^{1/4} = \sqrt{z}\sqrt{z_{1}} - \sqrt{z}(1 - x)^{1/4}\sqrt{z_{1}}(1 - x_{1})^{1/4}$$ and on simplification this gives $$(xx_{1})^{1/4} + \{(1 - x)(1 - x_{1})\}^{1/4} = 1$$ Putting $x = \alpha$, $x_{1} = \beta$ we get the modular equation as $$(\alpha\beta)^{1/4} + \{(1 - \alpha)(1 - \beta)\}^{1/4} = 1\tag{7}$$ In modern notation this is $$\sqrt{kl} + \sqrt{k'l'} = 1$$ which is the same as the equation derived using Jacobi's transformation of elliptic integrals.

Again from equation $(4)$ of current post and equations $(7), (8)$ of the previous post we immediately obtain $$\frac{\phi^{3}(q)}{\phi(q^{3})} + 2\cdot\frac{\phi^{3}(-q^{2})}{\phi(-q^{6})} = 3\phi(q)\phi(q^{3})\tag{8}$$ This can be transcribed to $$\frac{z^{3/2}}{z_{1}^{1/2}} + 2\cdot\frac{z^{3/2}(1 - x)^{3/8}}{z_{1}^{1/2}(1 - x_{1})^{1/8}} = 3z^{1/2}z_{1}^{1/2}$$ Ramanujan used the multiplier $m = z/z_{1} = K/L$ and then the above equation simplifies (on putting $x = \alpha, x_{1} = \beta$) to $$1 + 2\left(\frac{(1 - \alpha)^{3}}{1 - \beta}\right)^{1/8} = \frac{3}{m}\tag{9}$$ If we replace $\alpha$ by $(1 - \beta)$ and $\beta$ by $(1 - \alpha)$ the multiplier $m = K/L$ changes to $L'/K' = 3L/K = 3/m$ and hence we get $$1 + 2\left(\frac{\beta^{3}}{\alpha}\right)^{1/8} = m\tag{10}$$ From the previous post we have $$\left(\frac{\alpha^{3}}{\beta}\right)^{1/8} - \left(\frac{(1 - \alpha)^{3}}{1 - \beta}\right)^{1/8} = 1 = \left(\frac{(1 -\beta)^{3}}{1 - \alpha}\right)^{1/8} - \left(\frac{\beta^{3}}{\alpha}\right)^{1/8}\tag{11}$$ Hence we have the following relations $$\left(\frac{(1 - \alpha)^{3}}{1 - \beta}\right)^{1/8} = \frac{3 - m}{2m},\,\, \left(\frac{\beta^{3}}{\alpha}\right)^{1/8} = \frac{m - 1}{2}$$ $$\left(\frac{\alpha^{3}}{\beta}\right)^{1/8} = \frac{3 + m}{2m},\,\, \left(\frac{(1 -\beta)^{3}}{1 - \alpha}\right)^{1/8} = \frac{m + 1}{2}$$ Hence $$\alpha = \left(\frac{\alpha^{3}}{\beta}\right)^{3/8}\left(\frac{\beta^{3}}{\alpha}\right)^{1/8} = \frac{(3 + m)^{3}(m - 1)}{16m^{3}}$$ $$\beta = \left(\frac{\beta^{3}}{\alpha}\right)^{3/8}\left(\frac{\alpha^{3}}{\beta}\right)^{1/8} = \frac{(m - 1)^{3}(3 + m)}{16m}$$ so that $$m^{2}\alpha - \beta = (m^{2} - 1)\cdot\frac{3 + m}{2m} = (m^{2} - 1)\left(\frac{\alpha^{3}}{\beta}\right)^{1/8}$$ and finally we arrive at another form of the modular equation: $$m^{2} = \dfrac{\left(\dfrac{\alpha^{3}}{\beta}\right)^{1/8} - \beta}{\left(\dfrac{\alpha^{3}}{\beta}\right)^{1/8} - \alpha}\tag{12}$$

Modular Equations of Degree 5

From formula $(9)$ of this post we have $$\phi^{2}(q) = 1 + 4\left(\frac{q}{1 - q} - \frac{q^{3}}{1 - q^{3}} + \frac{q^{5}}{1 - q^{5}} - \frac{q^{7}}{1 - q^{7}} + \cdots\right)$$ and therefore $$\begin{align}\phi^{2}(q) - \phi^{2}(q^{5}) &= 4\left(\frac{q}{1 - q} - \frac{q^{3}}{1 - q^{3}} - \frac{q^{7}}{1 - q^{7}} + \frac{q^{9}}{1 - q^{9}}\right.\notag\\ &\left.\,\,\,\,\,-\, \frac{q^{11}}{1 - q^{11}} + \frac{q^{13}}{1 - q^{13}} + \frac{q^{17}}{1 - q^{17}} - \frac{q^{19}}{1 - q^{19}} + \cdots\right)\notag\\ &= 4\left(\frac{q + q^{9}}{1 + q^{10}} + \frac{q^{2} + q^{18}}{1 + q^{20}} + \frac{q^{3} + q^{27}}{1 + q^{30}} + \cdots\right)\notag\\ &\,\,\,\,\, -\, 4\left(\frac{q^{3} + q^{7}}{1 + q^{10}} + \frac{q^{6} + q^{14}}{1 + q^{20}} + \frac{q^{9} + q^{21}}{1 + q^{30}} + \cdots\right)\notag\\ &= 2\phi^{2}(-q^{10})\left(\frac{f(q, q^{9})}{f(-q, -q^{9})} - \frac{f(q^{3}, q^{7})}{f(-q^{3}, -q^{7})}\right)\notag\\ &= 2\phi^{2}(-q^{10})\left(\frac{f(q, q^{9})f(-q^{3}, -q^{7}) - f(-q, -q^{9})f(q^{3}, q^{7})}{f(-q, -q^{9})f(-q^{3}, -q^{7})}\right)\notag\\ &= 4q\phi^{2}(-q^{10})\frac{f(-q^{6}, -q^{14})f(-q^{2}, -q^{18})}{f(-q, -q^{9})f(-q^{3}, -q^{7})}\notag\\ &= 4qf(q, q^{9})f(q^{3}, q^{7})\notag\end{align}$$ In the last line above we used the relation $f(a, b)f(-a, b) = \phi(-ab)f(-a^{2}, -b^{2})$ and in the second last line we used conditional identitites for $f(a, b)$.

Now $$\begin{align} f(q, q^{9})f(q^{3}, q^{7}) &= (-q;q^{10})_{\infty}(-q^{9};q^{10})_{\infty}(-q^{3};q^{10})_{\infty}(-q^{7};q^{10})_{\infty}(q^{10};q^{10})_{\infty}^{2}\notag\\ &= \frac{(-q;q^{2})_{\infty}(q^{10};q^{10})_{\infty}^{2}}{(-q^{5};q^{10})_{\infty}}\notag\\ &= \chi(q)(q^{10};q^{10})_{\infty}^{2}(q^{5};q^{10})_{\infty}(-q^{10};q^{10})_{\infty}\notag\\ &= \chi(q)(q^{20};q^{20})_{\infty}(q^{5};q^{5})_{\infty}\notag\\ &= \chi(q)f(-q^{5})f(-q^{20})\notag\end{align}$$ Hence we have the following identity: $$\phi^{2}(q) - \phi^{2}(q^{5}) = 4q\chi(q)f(-q^{5})f(-q^{20})\tag{13}$$ Using transcription formulas with $x, y, z$ associated with $q$ and $x_{1}, y_{1}, z_{1}$ associated with $q^{5}$ we get $$\begin{align} z - z_{1} &= 4e^{-y}\cdot 2^{1/6}\{x(1 - x)e^{y}\}^{-1/24}\notag\\ &\,\,\,\,\,\,\,\times\, 2^{-1/6}\sqrt{z_{1}}(1 - x_{1})^{1/6}(x_{1}e^{y_{1}})^{1/24}\notag\\ &\,\,\,\,\,\,\,\times\, 4^{-1/3}\sqrt{z_{1}}(1 - x_{1})^{1/24}(x_{1}e^{y_{1}})^{1/6}\notag\end{align}$$ i.e. $$\frac{z}{z_{1}} - 1 = 2^{4/3}\frac{\{x_{1}(1 - x_{1})\}^{5/24}}{\{x(1 - x)\}^{1/24}}$$ Switching to $\alpha, \beta, m$ we get $$m = 1 + 2^{4/3}\left(\frac{\beta^{5}(1 - \beta)^{5}}{\alpha(1 - \alpha)}\right)^{1/24}\tag{14}$$ Replacing $m$ by $5/m$, $\alpha$ by $(1 - \beta)$, $\beta$ by $(1 - \alpha)$ we get another modular equation $$\frac{5}{m} = 1 + 2^{4/3}\left(\frac{\alpha^{5}(1 - \alpha)^{5}}{\beta(1 - \beta)}\right)^{1/24}\tag{15}$$ Again replacing $q$ by $(-q)$ in $(13)$ and on dividing the resulting equation by equation $(13)$ we get $$\frac{\phi^{2}(-q) - \phi^{2}(-q^{5})}{\phi^{2}(q) - \phi^{2}(q^{5})} = -\frac{\chi(-q)f(q^{5})}{\chi(q)f(-q^{5})}$$ Upon transcription the formula reduces to the following modular equation $$m = \dfrac{1 + \left(\dfrac{(1 - \beta)^{5}}{1 - \alpha}\right)^{1/8}}{1 + \{(1 - \alpha)^{3}(1 - \beta)\}^{1/8}}\tag{16}$$ Replacing $m$ by $5/m$, $\alpha$ by $(1 - \beta)$, $\beta$ by $(1 - \alpha)$ we get $$\frac{5}{m} = \dfrac{1 + \left(\dfrac{\alpha^{5}}{\beta}\right)^{1/8}}{1 + \{\alpha\beta^{3}\}^{1/8}}\tag{17}$$ To keep the current post to a manageable length, we will postpone the study of a few more modular equations of degree $5$ to the next post.

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