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Quiz: Convolutions
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last edited
by RH 12 years, 1 month ago
 Let and be continuously differentiable periodic functions. The derivative of the convolution is given by

 CORRECT. This is one of two correct answers.

 INCORRECT. You're thinking of the product rule: . Convolutions behave differently from products.

 INCORRECT. You're thinking of the sum rule: . Convolutions behave differently from sums.

 INCORRECT. You might be thinking of Abel's formula for matrices: . Convolutions behave differently from inverses.

 CORRECT. This is one of two correct answers.
 In general, there is no simple formula available.
 Let and be continuously differentiable periodic functions, and let be an integer. The Fourier coefficient of the convolution is given by

 In general, there is no simple formula available.
 Let and be continuously differentiable periodic functions. The average value of is equal to
 The difference between the average value of and the average value of .
 The average of the average value of and the average value of .
 The convolution of the average value of and the average value of .
 PARTIALLY. This is true if one thinks of the average values of and as constant functions rather than numbers, but this is a rather clumsy way to phrase the answer.
 The product of the average value of and the average value of .
 The sum of the average value of and the average value of .
 In general, there is no simple formula available.
 Let , , be continuous periodic functions. The expression can also be written as

 None of the above.
 Let , , be continuous periodic functions. The expression can also be written as

 None of the above.
 Let , , be continuous periodic functions. The expression can also be written as
 None of the above.
 CORRECT. In general, there is no useful formula for pulling a product out of a convolution (or a convolution out of a product).
 Let , be periodic functions. If is continuously differentiable, and is twice continuously differentiable, then the best we can say about is that it is periodic and
 Riemann integrable.
 Piecewise continuous.
 Continuous.
 Continuously differentiable.
 Twice continuously differentiable.
 Three times continuously differentiable.
 CORRECT. Convolving two functions combines their orders of smoothness together.
 Infinitely differentiable.
 Let , be periodic functions. If is continuously differentiable, and is twice continuously differentiable, then the best we can say about is that it is periodic and
 Riemann integrable.
 Piecewise continuous.
 Continuous.
 Continously differentiable.
 CORRECT. In general, the sum of two functions is only as smooth as the rougher of its two factors.
 Twice continuously differentiable.
 Three times continuously differentiable.
 Infinitely differentiable.
 Let , be periodic functions. If is continuously differentiable, and is twice continuously differentiable, then the best we can say about is that it is periodic and
 Riemann integrable.
 Piecewise continuous.
 Continuous.
 Continously differentiable.
 CORRECT. In general, the product of two functions is only as smooth as the rougher of its two factors.
 Twice continuously differentiable.
 Three times continuously differentiable.
 Infinitely differentiable.
 Let , be periodic functions. If and are Riemann integrable, then the best we can say about is that it is periodic and
 Bounded.
 Riemann integrable.
 PARTIALLY. While this true, more can be said.
 Piecewise continuous.
 Continuous.
 Continously differentiable.
 Twice continuously differentiable.
 Infinitely differentiable.
 Let , be periodic functions. If and are Riemann integrable, then the best we can say about is that it is periodic and
 Bounded.
 INCORRECT. While this true, more can be said.
 Riemann integrable.
 Piecewise continuous.
 Continuous.
 Continously differentiable.
 Twice continuously differentiable.
 Infinitely differentiable.
 Let be a periodic function, and let be the constant function . Then is
 The same function as .
 The constant function .
 The constant function with value equal to the mean of .
 The constant function with value equal to .
 The value of at the point .
 .
 Let be a continuous periodic function, and let be a family of approximations to the identity (a.k.a. good kernels). Which of the following statements is true?
 For each , converges to as goes to infinity.
 For each , converges to as goes to infinity.
 For each , converges to as goes to infinity.
 For each and each , we have .
 For each , converges to as goes to infinity.
 The functions converge to zero as goes to infinity.
Score:
.
Quiz: Convolutions

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Comments (1)
Mohammad said
at 1:40 am on Dec 29, 2010
Have anyone checked the "PDF version"?
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