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How this equation:
I_{s}=\sum_{b=1}^{Nu}w_{b}\int_{\infty }^{t}d\tau K_{s}(t\tau)u_{b}(\tau)
differentiates to this:
\tau_{s}\frac{\mathrm{d} I_{s} }{\mathrm{d} t}=I_{s}+\sum_{b=1}^{Nu}w_{b}u_{b}
given that Ks(t)=exp(tau/t)/tau
 8 months ago
 8 months ago
How this equation: I_{s}=\sum_{b=1}^{Nu}w_{b}\int_{\infty }^{t}d\tau K_{s}(t\tau)u_{b}(\tau) differentiates to this: \tau_{s}\frac{\mathrm{d} I_{s} }{\mathrm{d} t}=I_{s}+\sum_{b=1}^{Nu}w_{b}u_{b} given that Ks(t)=exp(tau/t)/tau
 8 months ago
 8 months ago

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amnsbrBest ResponseYou've already chosen the best response.0
\[I_{s}=\sum_{b=1}^{Nu}w_{b}\int\limits_{\infty }^{t}d\tau K_{s}(t\tau)u_{b}(\tau)\] \[\tau_{s}\frac{\mathrm{d} I_{s} }{\mathrm{d} t}=I_{s}+\sum_{b=1}^{Nu}w_{b}u_{b}\]
 8 months ago

SithsAndGigglesBest ResponseYou've already chosen the best response.0
Let me just rewrite this question so it's easier to read: \[\large I_s=\sum_{b=1}^\nu w_b\int_{\infty}^t\frac{\exp\left[\frac{\tau}{t\tau}\right]}{\tau}u_b(\tau)~d\tau\] And you're asking how the derivative (with respect to t, apparently) is \[\large \tau_s\frac{dI_s}{dt}=I_s+\sum_{b=1}^\nu w_bu_b\] Am we to assume \(\large u_b\) is the Heaviside step function?
 8 months ago

amnsbrBest ResponseYou've already chosen the best response.0
ub() is convolution of pb() with a boxcar filter and pb() is a sum of dirac functions i don't know what is Heaviside step function, sry!
 8 months ago

SithsAndGigglesBest ResponseYou've already chosen the best response.0
I don't remember much about convolution, and I've never heard of the boxcar filter you mention. Sorry I can't help! By the way, the Heaviside step function is \[u_t(x)=u(xt)=\begin{cases}1&\text{for }x\ge t\\0&\text{for }x<t\end{cases}\] You might know it by some other name.
 8 months ago

amnsbrBest ResponseYou've already chosen the best response.0
u(t) is the firing rate of a neuron and is continous in time,i think you can assume that it's just an arbitrary function of time! About Heaviside,yes it is similar to Delta dirac function btw, boxcar filter is a simple filter which is 1 in [a,b] and zero otherwise
 8 months ago

SithsAndGigglesBest ResponseYou've already chosen the best response.0
@phi, it's the given as exponential function in the integral. \[\large K_s(t)=\frac{\exp\left[\frac{\tau}{t}\right]}{\tau}\]
 8 months ago

phiBest ResponseYou've already chosen the best response.0
Do we know anything about \( w_b(t)\) ? in the meantime, I think we can use this http://mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html
 8 months ago
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