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Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing $k$ so long as $2x^2(k+1)(N-1)<(N-2k-1)(N-2k)$. Solving for $k$ (and making approximations valid for large $N$) gives $$\begin{align}k &< \frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\\\ &< \frac{N x^2+ 2N-\sqrt{x^4+x^2}N}{4}=\frac{2+x^2-\sqrt{x^4+x^2}}{4}N\\\\ \end{align}$$ and so for large $N$ the largest term will be $k=f(x)N$, with $f(x)=\frac{2+x^2-\sqrt{x^4+x^2}}{4}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any $\epsilon$ we have arbitrarily large $N$ such that this is within $\epsilon$ of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y$) we get $$\begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\\\ \end{align}$$ and so (note we make the approximation $\log(N-1)\approx \log(N)$) $$\begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\\\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\\\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ \end{align}$$ thus $$\begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\\\ \end{align}$$ which should be quite close to the actual value due to the accuracy of Stirling's for large values.

Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing $k$ so long as $2x^2(k+1)(N-1)<(N-2k-1)(N-2k)$. Solving for $k$ (and making approximations valid for large $N$) gives $$\begin{align}k &< \frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\\\ &< \frac{N x^2+ 2N-\sqrt{x^4+4x^2}N}{4}=\frac{2+x^2-\sqrt{x^4+4x^2}}{4}N\\\\ \end{align}$$ and so for large $N$ the largest term will be $k=f(x)N$, with $f(x)=\frac{2+x^2-\sqrt{x^4+4x^2}}{4}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any $\epsilon$ we have arbitrarily large $N$ such that this is within $\epsilon$ of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y$) we get $$\begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\\\ \end{align}$$ and so (note we make the approximation $\log(N-1)\approx \log(N)$) $$\begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\\\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\\\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ \end{align}$$ thus $$\begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\\\ \end{align}$$ which should be quite close to the actual value due to the accuracy of Stirling's for large values.

Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing $k$ so long as $2x^2(k+1)(N-1)<(N-2k-1)(N-2k)$. Solving for $k$ (and making approximations valid for large $N$) gives $$\begin{align}k &< \frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\\\ &< \frac{N x^2+ 2N-\sqrt{x^4+x^2}N}{4}=\frac{1+x^2-\sqrt{x^4+x^2}}{4}N\\\\ \end{align}$$ and so for large $N$ the largest term will be $k=f(x)N$, with $f(x)=\frac{1+x^2-\sqrt{x^4+x^2}}{4}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any $\epsilon$ we have arbitrarily large $N$ such that this is within $\epsilon$ of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y$) we get $$\begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\\\ \end{align}$$ and so (note we make the approximation $\log(N-1)\approx \log(N)$) $$\begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\\\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\\\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}+f(x)-2-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ \end{align}$$ thus $$\begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\\\ \end{align}$$ which should be quite close to the actual value due to the accuracy of Stirling's for large values.

Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing $k$ so long as $2x^2(k+1)(N-1)<(N-2k-1)(N-2k)$. Solving for $k$ (and making approximations valid for large $N$) gives $$\begin{align}k &< \frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\\\ &< \frac{N x^2+ 2N-\sqrt{x^4+x^2}N}{4}=\frac{2+x^2-\sqrt{x^4+x^2}}{4}N\\\\ \end{align}$$ and so for large $N$ the largest term will be $k=f(x)N$, with $f(x)=\frac{2+x^2-\sqrt{x^4+x^2}}{4}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any $\epsilon$ we have arbitrarily large $N$ such that this is within $\epsilon$ of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y$) we get $$\begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\\\ \end{align}$$ and so (note we make the approximation $\log(N-1)\approx \log(N)$) $$\begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\\\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\\\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ \end{align}$$ thus $$\begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\\\ \end{align}$$ which should be quite close to the actual value due to the accuracy of Stirling's for large values.

Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k-2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{(k+1)(N-1)}{2x^2(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing $k$ so long as $(k+1)(N-1)< 2x^2(N-2k-1)(N-2k)$. Solving for $k$ (and making approximations valid for large $N$) gives $$\begin{align}k &< \frac{-1+N-4 x^2+8 N x^2-\sqrt{1-2 N+N^2-24 x^2+8 N x^2+16 N^2 x^2+16 x^4}}{16 x^2}\\\\ &< \frac{N+8 N x^2-\sqrt{N^2+16 N^2 x^2}}{16 x^2}=\frac{1+8x^2-\sqrt{1+16x^2}}{16 x^2}N\\\\ \end{align}$$ and so for large $N$ the largest term will be $k=f(x)N$, with $f(x)=\frac{1+8x^2-\sqrt{1+16x^2}}{16 x^2}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any $\epsilon$ we have arbitrarily large $N$ such that this is within $\epsilon$ of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y$) we get $$\begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\\\ \end{align}$$ and so (note we make the approximation $\log(N-1)\approx \log(N)$) $$\begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\\\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\\\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}+f(x)-2-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ \end{align}$$ thus $$\begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\\\ \end{align}$$ which should be quite close to the actual value due to the accuracy of Stirling's for large values.

Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing $k$ so long as $2x^2(k+1)(N-1)<(N-2k-1)(N-2k)$. Solving for $k$ (and making approximations valid for large $N$) gives $$\begin{align}k &< \frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\\\ &< \frac{N x^2+ 2N-\sqrt{x^4+x^2}N}{4}=\frac{1+x^2-\sqrt{x^4+x^2}}{4}N\\\\ \end{align}$$ and so for large $N$ the largest term will be $k=f(x)N$, with $f(x)=\frac{1+x^2-\sqrt{x^4+x^2}}{4}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any $\epsilon$ we have arbitrarily large $N$ such that this is within $\epsilon$ of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y$) we get $$\begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\\\ \end{align}$$ and so (note we make the approximation $\log(N-1)\approx \log(N)$) $$\begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\\\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\\\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ &\approx -\frac{\log(2)}{2}+f(x)-2-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\\\ \end{align}$$ thus $$\begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\\\ \end{align}$$ which should be quite close to the actual value due to the accuracy of Stirling's for large values.