Derivate of generalized function

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The discussion revolves around demonstrating the relationship between the derivative of a generalized function and integration by parts. The user seeks clarification on the appearance of a negative sign in the equation involving the integrals of the generalized functions. They derive the equation using limits and integration by parts, concluding that the integral of the derivative of a function multiplied by another function equals the negative integral of the original function multiplied by the derivative of the second function. The confusion arises from the application of integration by parts, where the boundary terms vanish under certain conditions. Overall, the discussion emphasizes the importance of understanding the properties of generalized functions in this context.
greisen
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Hi,

I have to show that if the derivate f'(x) of a generalized function f(x) is defined by the sequence f'_n(x) where f(x) is defined

f_n(x)[\tex] <br /> <br /> then <br /> <br /> \int_{-\infty}^{\infty}f&amp;#039;(x)F(x) dx = - \int_{-\infty}^{\infty}f(x)F&amp;#039;(x) dx<br /> <br /> I use the limits for generalized functions and get<br /> <br /> lim_{n \to \infty} \int_{-\infty}^{\infty}f&amp;#039;_n(x)F(x) dx = - \int_{-\infty}^{\infty}f_n(x)F&amp;#039;(x) dx<br /> <br /> which should show the above - I am a liltte confused where the minus sign comes from?<br /> - \int_{-\infty}^{\infty}f(x)F&amp;#039;(x) dx<br /> <br /> Any help appreciated - thanks in advance
 
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it is from integration by parts
let f,F be functions
d(fF)=(fF)'dx=f'Fdx+fF'dx
f'Fdx=d(fF)-fF'dx
integrating gives (assume f,F->0)
ʃf'Fdx=ʃd(fF)-ʃfF'dx
ʃd(fF)=0 so
ʃf'Fdx=-ʃfF'dx
now extent this to generalized functions by limits
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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