University:
Stanford University
Course:
MATH 205B | Real Analysis
Academic year:

2023
Views:

30

Pages:

6
Author:

MathPaladin

Questions and Answers Sheet 6 Analysis Question #1 1 Asymptotic behaviour of ∫0 𝑓(𝑥)𝑥 𝑛 𝑑𝑥 as 𝑛 → ∞ Answer: Since f is continuous on a closed bounded interval, it has a maximum absolute value 𝑀 ≥ 0 on that interval, so you have 1 1 1 |∫ 𝑓(𝑥)𝑥 𝑛 𝑑𝑥| ≤ ∫ |𝑓(𝑥)𝑥 𝑛 |𝑑𝑥 ≤ 𝑀 ∫ 𝑥 𝑛 𝑑𝑥 0 0 𝑀 = . 𝑛+1 0 Question #2 Comparison of 𝐿2 and 𝐿1 norms for functions In the context of Lebesgue Integrals, I have came across 𝐿2 as the set of measurable functions 𝑓: [𝑎, 𝑏] → 𝐶 that have Lebesgue integrable squares - that is 𝑥 → |𝑓(𝑥)|2 is Lebesgue integrable. Answer: Yes, the inequality is given by Cauchy-Schwarz: b b b a a a || f ||L1 =  | f ( x) | dx  (  | f ( x) |2 dx)1/2 (  dx)1/2 = b − a || f || L2 Question #3 Prove that the equation 2𝑥 − 1 = 𝑦 2 has no integer solutions where x>1 . Answer: 𝑦 = 2𝑥 − 1 has a root for x=0 , because its asymptotic at y=-1 and is continually increasing. However, 𝑦 2 = 2𝑥 − 1 tells a different story. It still has a root at x=0 , but its parabolic in nature in the horizontal direction. 𝑑𝑦 The reason your function has no roots because of your domain of 𝑥 > 1 ⋯ > 0, for x>1 and at x=1, 𝑑𝑥 y=1. Hence, what can you conclude about the nature of the function and its position in a cartesian plane after x>1 . Will it every hit the x-axis? Question #4 √𝑛 Why lim (22 ) 𝑛 𝑛→∞ (1+(10)−22 ) Answer: Observe that for n big enough we have 5 < (1 + 10−4 )√𝑛 So 5√𝑛 < (1 + 10−4 )𝑛 4 √𝑛 and then < (1+10−4 )𝑛 4 √𝑛 5√𝑛 4 √𝑛 =( ) 5 →0 Question #5 How can we verify this limit via 𝜖 − 𝛿 method? I have to calculate the limit lim 1−cos(𝑥) 𝑥→0 3 sin2 (𝑥) Answer: We have to show that 𝜋 ∀𝜖 > 0: ∃𝛿 ∈ (0, ] : ∀𝑥 ∈ 𝑅 {0} 2 : (|𝑥| < 𝛿 ⇒ |𝑓(𝑥)| < 𝜖) (1−cos 𝑥)2 where 𝑓(𝑥) = = 6 sin2 𝑥 (1−cos 𝑥)2 6(1−cos2 𝑥) Let 𝑡 = cos 𝑥 . Then, because 𝑡 ≠ 1 whenever 𝑥 ≠ 0 and |𝑥| < (1 − 𝑡)2 (1 − 𝑡) (2 − 1 − 𝑡) 𝑓(𝑥) = = = 2 6(1 − 𝑡 ) 6(1 + 𝑡) 6(1 + 𝑡) 1 1 = − 3(1+𝑡) 𝜋 2 , 6 f(x) strictly decreases w.r.t.t. 2 Find t that makes 𝑓(𝑥) = 𝜖. After all, 𝑡 = (6𝜖+1) − 1. Since 𝑡 = cos 𝑥 , choosing 𝛿 = 𝑎𝑟𝑐𝑐𝑜𝑠 ( 2 1 6 6𝑚𝑖𝑛{𝜖, }+1 − 1) proves the claim. I think you can show that 𝛿 is an increasing function of 𝜖. Question #6 Sum of bounds of ∑ 1 𝑎𝑛 +𝑏 𝑛 Answer: We can suppose b 𝑥 2 } is open with open set definition Answer: Hint: 𝑧 − 𝑤 2 = (𝑥0 − 𝑦02 ) + (𝑧 − 𝑥0 ) + (𝑦02 − 𝑤 2 ) > (𝑥0 − 𝑦02 ) − |𝑧 − 𝑥0 | − |𝑤 − 𝑦0 ||𝑤 + 𝑦0 | ≥ (𝑥0 − 𝑦02 ) − |𝑧 − 𝑥0 | − |𝑤 − 𝑦0 |(2𝑦0 − |𝑤 − 𝑦0 |) > (𝑥0 − 𝑦02 ) − 𝜖 − 𝜖(2𝑦0 − 𝜖) (Note that 𝑦0 > 0 ). Question #13 1 1 Proof that 𝑎𝑚 < 𝑎𝑛 with 𝑚 < 𝑛, 0 < 𝑎 < 1 and 𝑛, 𝑚 ∈ 𝑁 Answer: Rise both side of the inequality to the power m\times n to find 𝑎𝑛 < 𝑎𝑚 that is true since so since n>m 𝑎𝑛 = 𝑎𝑚 × 𝑎𝑛−𝑚 < 𝑎𝑚 because 𝑎𝑛−𝑚 < 1 Question #14 |𝑓(𝑍𝑝)| has at most 0 as a limit point in the non-zero reals Answer: The key concept is this: the non-zero accumulation points are deceptive, in the sense that if 𝑥𝑛 → 𝑥 in 𝐶𝑝 , then either |𝑥𝑛 | is eventually constant, or |𝑥𝑛 | → 0 ; in other words, if |𝑥𝑛 | tends to say 1, and |𝑥𝑛 | is strictly increasing, then there is no chance that 𝑥𝑛 actually converges. With this in mind, it is now a trivial application of the ''Bolzano–Weierstrass theorem'': Assume 𝑦 ∈ 𝑅 is a non-zero accumulation point, so there are 𝑥𝑛 ∈ 𝑍𝑝 such that |𝑓(𝑥𝑛 )| → 𝑦 and |𝑓(𝑥𝑛 )| are all distinct. By ''Bolzano–Weierstrass theorem'', 𝑥𝑛 has a convergent subsequence, so we might WLOG assume that 𝑥𝑛 → 𝑥 ; since 𝑓(𝑥𝑛 ) → 𝑓(𝑥) and |𝑓(𝑥𝑛 )| are all distinct, we must have |𝑓(𝑥𝑛 )| = 𝑦 = 0 , contradiction. Question #15 Prove that 𝑓(𝑥) = 𝑥 2 is integrable using superior and inferior sums Answer: The 𝑃1 is a partion on [-1,0] and your 𝑃2 is a partition on [0,2] . So your partition P on [-1,2] should be 𝑃 = 𝑃1 + 𝑃2 aka: 1 1 2 𝑃 = {𝑡−𝑛 = −1, 𝑡−𝑛+1 = −1 + , ⋯ , 𝑡−1 = − , 𝑡0 = 0, 𝑡1 = ⋯ , 𝑡𝑛 = 2} 𝑛 𝑛 𝑛 You can then start from here to show 𝑆(𝑓, 𝑃) − 𝑠(𝑓, 𝑃) < 𝜖, or [𝑆(𝑓, 𝑝1 ) + 𝑆(𝑓, 𝑝2 )] − [𝑠(𝑓, 𝑝1 ) + 𝑠(𝑓, 𝑝2 )] < 𝜖. Question #16 Prove inequality 𝑒 𝑥 > 𝑥 𝑛+1 𝑛+1! Answer: You can use Taylor series of the exponential, whose radius of convergence is infinite. 𝑒 𝑥 = ∑+∞ 𝑘=0 𝑥𝑘 𝑘! Can you proceed? Observe that due to the previous series, we can write the partial sum for k from 0 to n+2 : 𝑥2 𝑥𝑛 𝑥 𝑛+1 𝑥 𝑛+2 1 +𝑥 + + ⋯+ + + +⋯ 2! 𝑛! (𝑛 + 1)! (𝑛 + 2)! And this, being a sum of positive terms, is surely greater than 𝑥 𝑛+1 (𝑛+1)! Question #17 A differentiable form on an open subset U subseteq 𝑅𝟚 is a map 𝜔 = 𝜔𝑥 𝑑𝑥 + 𝜔𝑦 𝑑𝑦 for some 𝜔𝑥 , 𝜔𝑦 ∈ 𝐶 1 (𝑈) For example, if 𝑓 ∈ 𝐶 1 (𝑈), 𝜔 = 𝑑𝑓 = 𝜕𝑓 𝜕𝑥 𝑑𝑥 + 𝜕𝑓 𝜕𝑦 𝑑𝑦 is a differential form. My question is: what is the difference between 𝜕𝑥 and dx? Answer: dx is a basis vector in the space of 1-forms for functions in C1. On the other hand, ∂x is not a symbol with any meaning: it does denote any particular mathematical object. The symbol ∂x is merely an artifact of notation: ∂f∂x is merely how we choose to denote the partial derivative of f with respect to x. We could choose to denote it differently, avoiding the symbol ∂x altogether, since it has no mathematical significance. Why was this symbol chosen? I am sure there is some history behind the symbol, though I am not knowledgeable in it. But one could very well choose to denote it as fx instead, and now you communicate the same idea, without appealing to strange notation. Question #18 With cash =$2950 and accounts payable at $11,900, what's the cash ratio? Answer: This is an accounting question - not sure where it should go, so I'll keep it here in Stats. The cash ratio is the ratio of cash to current liabilities (we're in essence seeing how many times we can pay current liabilities with the cash on hand). We have: Cash = 2950 𝐴 = 11900 𝑃 and so: Cash ratio = 2950 11900 ≅ 24.8% Question #19 Point R is on line segment QR. Given QR=3x-8, QS=3x, and RS=2x, determine the numerical lenght of RS. Answer: Given that R is on QS, then QR + RS = QS, substitute values 3x - 8 + 2x = 3x, that is 5x - 8 = 3x (subtract 3x from both sides) 2x - 8 = 0 (add 8 to both sides) 2x = 8 (divide both sides by 2) x=4 Thus RS = 2x = 2 × 4 = 8