Question:
Estimate the value of $ \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}f(x)}dx} $ using the saddle point approximation, with $ \epsilon = 0.05 $ and $ f (x) = \frac{1}{2}x^4 + x + 1 $.
Solution:
Let's take a look at the derivatives of $ f(x) $, we have
$ \begin{eqnarray*} \frac{d}{dx} f(x) &=& 2x^3 + 1 \\ \frac{d^2}{dx^2} f(x) &=& 6x^2 \\ \end{eqnarray*} $
Now we see the function is convex, we also see $ m = -\frac{1}{\sqrt[3]{2}} $ is the global minimum of the function. The power in the exponential is negative, which means for large numbers, the value of the exponential is essentially 0, and therefore we can concentrate near the global minimum. Let's expand $ f(x) $ as a Taylor series around the global minimum, we have:
$ \begin{eqnarray*} f(x) &=& f(m) + \frac{f''(m)(x-m)^2}{2} + ... \end{eqnarray*} $
Ignoring the other coefficients we have got a Gaussian integral, that can be computed as:
$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}f(x)}dx} & \approx & \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}(f(m) + \frac{f''(m)(x-m)^2}{2})}dx} \\ & = & \frac{\sqrt{\pi}e^{-\frac{1}{\epsilon} f(m)}}{\sqrt{\frac{1}{2\epsilon} f''(m)}} \\ \end{eqnarray*} $
The numeric value of the above expression is $ 8.799 \times 10^-5 $, which matches quite well with a numerical integration, so that's the answer! To confess - I made a mistake once forgetting the 2 comes from the Taylor series, but thanks Coursera for allowing a second attempt.
Estimate the value of $ \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}f(x)}dx} $ using the saddle point approximation, with $ \epsilon = 0.05 $ and $ f (x) = \frac{1}{2}x^4 + x + 1 $.
Solution:
Let's take a look at the derivatives of $ f(x) $, we have
$ \begin{eqnarray*} \frac{d}{dx} f(x) &=& 2x^3 + 1 \\ \frac{d^2}{dx^2} f(x) &=& 6x^2 \\ \end{eqnarray*} $
Now we see the function is convex, we also see $ m = -\frac{1}{\sqrt[3]{2}} $ is the global minimum of the function. The power in the exponential is negative, which means for large numbers, the value of the exponential is essentially 0, and therefore we can concentrate near the global minimum. Let's expand $ f(x) $ as a Taylor series around the global minimum, we have:
$ \begin{eqnarray*} f(x) &=& f(m) + \frac{f''(m)(x-m)^2}{2} + ... \end{eqnarray*} $
Ignoring the other coefficients we have got a Gaussian integral, that can be computed as:
$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}f(x)}dx} & \approx & \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}(f(m) + \frac{f''(m)(x-m)^2}{2})}dx} \\ & = & \frac{\sqrt{\pi}e^{-\frac{1}{\epsilon} f(m)}}{\sqrt{\frac{1}{2\epsilon} f''(m)}} \\ \end{eqnarray*} $
The numeric value of the above expression is $ 8.799 \times 10^-5 $, which matches quite well with a numerical integration, so that's the answer! To confess - I made a mistake once forgetting the 2 comes from the Taylor series, but thanks Coursera for allowing a second attempt.
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