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Can we solve for $g$ when $\varepsilon$ is small?

$\newcommand{\sinc}{\operatorname{sinc}}$ $$3\sinc\left(-1+ \frac\varepsilon T \right)-3\sinc\left(1+\frac\varepsilon T\right)-\sinc\left(-3+\frac\varepsilon T\right)+\sinc\left(3+\frac\varepsilon T\right) = \frac\varepsilon g$$

I am not able to get a constant value for $g$. Can one do this?

Michael Hardy
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Elnaz
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1 Answers1

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Because $\operatorname{sinc}$ is an even function, for $\epsilon=0$ the left side is zero. Is $\epsilon$ supposed to be small? If so, we can use $\dfrac d{dx} \operatorname{sinc}(x)=\dfrac {x \cos x-\sin x}{x^2}$ so you can evaluate the first-order Taylor expansion. The first term will contribute $3\dfrac \epsilon T (- \cos 1+\sin 1)$ as will the second. You get $6\dfrac \epsilon T (- \cos 1+\sin 1)+2\dfrac \epsilon {9T} (- 3\cos 3+\sin 3)=\dfrac \epsilon g$ The epsilons divide out and you can invert both sides to find $g$

Ross Millikan
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  • Thanks Michael, shouldn't it be -2/9T? Actually, I know the final answer for g should be 3/16 but I cannot get it!! – Elnaz Oct 08 '14 at 01:57
  • $g$ has to be proportional to $T$ because the left side is all $\frac \epsilon T$ and the right side just has $\epsilon$ You also have the trig functions, so it shouldn't be that simple. Please use parentheses when using the divide slash. Is the $T$ in the numerator or denominator? – Ross Millikan Oct 08 '14 at 02:21
  • Yes, g is 3T/16. And, the above answer with the correction of -2ϵ/(9T) and also using the normalized sinc instead of the regular one gives the correct g = 3T/16. – Elnaz Oct 08 '14 at 02:51
  • But what about the terms $\cos 1, \sin 1, \cos 3, \sin 3?$ – Ross Millikan Oct 08 '14 at 03:13
  • By using normalized sinc(x)=sin(pix)/(pix), those terms become cos(pi), sin(pi), etc. – Elnaz Oct 08 '14 at 14:37
  • Ok, if your sinc includes the factor pi. It doesn't always. – Ross Millikan Oct 08 '14 at 14:42