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How to prove that $(1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1+\alpha$1-\alpha x$ (with $\alpha \approx 1$) for $x\ll 1$ in this specific case

After multiple plots I noticed that function $h(x)= (1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1+\alpha x$$1-\alpha x$ (with $\alpha \approx 1$), for $x\ll 1$ (specifically $0<x<0.1$) and $a=(K-1)d$, $b=K$, c=$Kd$; with $K \gg 1$, $K \gg d$ and $d>1$.

In other words, I need to prove that the first derivative of $h(x)$ is constant ($\approx 1$$\approx -1$) for $0<x<0.1$, Or equivalently that :$-b(1-x)^{b-1} \, _2F_1(a,b;c;x) + \frac{ab}{c} (1-x)^b \, _2F_1(a+1,b+1;c+1;x) \approx \text{constant} (\approx 1)$$-b(1-x)^{b-1} \, _2F_1(a,b;c;x) + \frac{ab}{c} (1-x)^b \, _2F_1(a+1,b+1;c+1;x) \approx \text{constant} (\approx -1)$.

How to prove that $(1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1+\alpha x$ (with $\alpha \approx 1$) for $x\ll 1$ in this specific case

After multiple plots I noticed that function $h(x)= (1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1+\alpha x$ (with $\alpha \approx 1$), for $x\ll 1$ (specifically $0<x<0.1$) and $a=(K-1)d$, $b=K$, c=$Kd$; with $K \gg 1$, $K \gg d$ and $d>1$.

In other words, I need to prove that the first derivative of $h(x)$ is constant ($\approx 1$) for $0<x<0.1$, Or equivalently that :$-b(1-x)^{b-1} \, _2F_1(a,b;c;x) + \frac{ab}{c} (1-x)^b \, _2F_1(a+1,b+1;c+1;x) \approx \text{constant} (\approx 1)$.

How to prove that $(1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1-\alpha x$ (with $\alpha \approx 1$) for $x\ll 1$ in this specific case

After multiple plots I noticed that function $h(x)= (1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1-\alpha x$ (with $\alpha \approx 1$), for $x\ll 1$ (specifically $0<x<0.1$) and $a=(K-1)d$, $b=K$, c=$Kd$; with $K \gg 1$, $K \gg d$ and $d>1$.

In other words, I need to prove that the first derivative of $h(x)$ is constant ($\approx -1$) for $0<x<0.1$, Or equivalently that :$-b(1-x)^{b-1} \, _2F_1(a,b;c;x) + \frac{ab}{c} (1-x)^b \, _2F_1(a+1,b+1;c+1;x) \approx \text{constant} (\approx -1)$.

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tam
tam
  • 233
  • 2
  • 7

How to prove that $(1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1+\alpha x$ (with $\alpha \approx 1$) for $x\ll 1$ in this specific case

After multiple plots I noticed that function $h(x)= (1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1+\alpha x$ (with $\alpha \approx 1$), for $x\ll 1$ (specifically $0<x<0.1$) and $a=(K-1)d$, $b=K$, c=$Kd$; with $K \gg 1$, $K \gg d$ and $d>1$.

In other words, I need to prove that the first derivative of $h(x)$ is constant ($\approx 1$) for $0<x<0.1$, Or equivalently that :$-b(1-x)^{b-1} \, _2F_1(a,b;c;x) + \frac{ab}{c} (1-x)^b \, _2F_1(a+1,b+1;c+1;x) \approx \text{constant} (\approx 1)$.