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SUGGESTIONS FOR ANALYSIS II - SERIES 16 Exercise 61 Let us consider a differential equation of the form 0 y = f (x)g(y) y(x0 ) = y0 To find a solution we use the general formula Z x Z y du f (s) d s = x0 y0 g(u) and we try to explicit y as function of x. Pay attention to the correct domain of definition of f and g. Exercise 62 • To draw a graphic of φ we just study the given function. To prove that φ is a C ∞ function, we notice that φ is of class C ∞ in the open intervals (−∞, 0) and (0, +∞). To check that φ is of class C ∞ also at t = 0 we just compute the derivatives of φ on the open intervals (−∞, 0) and (0, +∞) and then we take te limit in both sides as t → 0. To be more precise, for each k ≥ 0 we have φ(k) (t) = 0 (t) − 1 φ(k) (t) = ptk2k e t if t ∈ (−∞, 0) if t ∈ (0, +∞) where pk (t) is a polynomial in t with real coefficients. This implies that φ(k) (t) goes to 0 as t → 0+ . On the other hand, φ(k) (t) is constantly 0 for t < 0. So that, lim φ(k) (t) = 0 t→0 and this proves that φ is of class C k for each k ≥ 0. Thus, φ is of class C ∞ . • It is enough to set 1 ak = R +∞ φ(1 − k 2 x2 ) d x −∞ and noticing that supp(φk ) = {x ∈ R | φk (x) 6= 0} ⊆ [−k.k] Exercise 63 First of all we recall the following formula Z π 1 1 i(d−k)x (1) e dx = 0 2π −π if d = k if d = 6 k Thanks to this, for |k| ≤ n we find ! Z π Z π n X 1 1 −ikx idx fˆ(k) = f (x)e dx = cd e e−ikx d x = 2π −π 2π −π d=−n Z π n X 1 = cd ei(d−k)x d x = ck 2π −π d=−n and similarly we deduce that fˆ(k) = 0 for |k| > 0. 1 2 SUGGESTIONS FOR ANALYSIS II - SERIES 16 To prove the Parseval identity we recall (1) and the fact that ck if |k| ≤ n ˆ f (k) = 0 if |k| > n in order to write ! ! Z π Z π X X 1 1 |f (x)|2 d x = cd eidx ck eikx d x = 2π −π 2π −π d∈Z k∈Z Z π X X 1 cd eidx = ck eikx d x = 2π −π −n≤d≤n = X −n≤k≤n cd ck fˆ(d)fˆ(k) −n≤d,k≤n = X −n≤k≤n |ck |2 = X 1 2π Z π e i(d−k)x dx = −π |fˆ(k)|2 k∈Z Exercise 64 • From Newton binomial formula we now that n X n k n−k a b (a + b)n = k k=0 We use this hint and write 1 = x + (1 − x) to get the desired result. • Taking partial derivatives with respect to a in the Newton binomial formula we get n X n(a + b)n−1 = kak−1 bn−k k=0 and then, if we multiplicate by a both sides in the previous equation we find n X n−1 na(a + b) = kak bn−k k=0 So that, to prove the given statement it is enough to substitute a = x and b = 1 − x. For the second formula we just need to take the second partial derivative with respect to a in the Newton binomial formula and finally substitute a = x and b = 1 − x. • Look at Section 3 of the given side notes.