Questions and Answers #5 Analysis

Questions and Answers Sheet 5 Analysis Question #1 0 0 ∫1 tan(𝑥 2 ) 𝑑𝑥 + 2 ∫1 𝑥 2 sec 2 (𝑥 2 ) 𝑑𝑥 = Select one: True False 𝜋 4 Answer: Step1 This qestion belongs to the definite intergal in which we have to justify the given statement is true or false by providing the correct solution of the given question. We can write the given statement below 0 0 𝜋 ∫1 tan(𝑥 2 ) 𝑑𝑥 + 2 ∫1 𝑥 2 sec(𝑥 2 ) 𝑑𝑥 = to solve the question we move to the nest step-2 4 Step 2 first, we write the given statement below and take the LHS side of the given statement so 0 0 𝜋 ∫1 tan(𝑥 2 ) 𝑑𝑥 + 2 ∫1 𝑥 2 sec(𝑥 2 ) 𝑑𝑥 = 4 now take the LHS side 0 0 𝐿𝐻𝑆 = ∫ tan(𝑥 2 ) 𝑑𝑥 + 2 ∫ 𝑥 2 sec(𝑥 2 ) 𝑑𝑥 0 1 1 0 = ∫ tan(𝑥 2 ) 𝑑𝑥 + 2 ∫ 𝑥 2 sec(𝑥 2 ) 𝑑𝑥 1 1 Now use the integration by part consider p and q function of x 𝑑𝑝 ∫ 𝑝𝑞𝑑𝑥 = 𝑝 ∫ 𝑞𝑑𝑥 − ∫ (𝑑𝑥 ∫ 𝑞𝑑𝑥) 𝑑𝑥 now apply 0 0 = ∫1 tan(𝑥 2 ) 𝑑𝑥 + 2 ∫1 𝑥 2 sec(𝑥 2 ) 𝑑𝑥 0 0 = tan(𝑥 2 ) ∫1 1𝑑𝑥 − ∫1 ( 0 𝑑 = tan(𝑥 2 ) 𝑥|10 − ∫1 (sec(𝑥 0 0 0 𝑡𝑎𝑛(𝑥 2 ) ∫1 1𝑑𝑥 ) 𝑑𝑥 + 2 ∫1 𝑥 2 sec(𝑥 2 ) 𝑑𝑥 𝑑𝑥 2) ( 𝑑 𝑑𝑥 0 𝑥 2 ) . 𝑥) 𝑑𝑥 + 2 ∫1 𝑥 2 sec(𝑥 2 ) 𝑑𝑥 0 = tan(𝑥 2 ) 𝑥|10 − ∫1 (𝑠𝑒𝑐(𝑥 2 )(2𝑥). 𝑥)𝑑𝑥 + 2 ∫1 𝑥 2 𝑠𝑒𝑐(𝑥 2 )𝑑𝑥 0 0 = tan( x 2 ) x |10 −2  ( sec(x 2 )(2 x) x.x)) dx + 2  x 2 sec( x 2 ) dx 1 1 0 0 = tan(𝑥 2 ) 𝑥|10 − 2 ∫1 (𝑠𝑒𝑐(𝑥 2 )(𝑥 2 ))𝑑𝑥 + 2 ∫1 𝑥 2 𝑠𝑒𝑐(𝑥 2 )𝑑𝑥 = tan (𝑥^2)𝑥|_0^1 − 2∫ _{1}^{0}(𝑠𝑒𝑐(𝑥^2)𝑑𝑥 + 2∫ _{1}^{0}𝑥^2𝑠𝑒𝑐(𝑥^2)𝑑𝑥 = tan(𝑥 2 ) 𝑥|10 − 0 now put the limits = tan(𝑥 2 ) 𝑥|10 = tan(12 ) (1) − 𝑡𝑎𝑛(02 )(0) − 𝑡𝑎𝑛(1) − 0 𝐿𝐻𝑆 = 𝑡𝑎𝑛(1) 𝜋 Now we got the LHS value and the RHS value is from the statement 0 4 0 ∫ 𝑡𝑎𝑛(𝑥 2 )𝑑𝑥 2 + 2 ∫ 𝑥 𝑠𝑒𝑐(𝑥 1 1 2 )𝑑𝑥 𝜋 = 4 and 0 0 𝐿𝐻𝑆 = ∫1 𝑡𝑎𝑛(𝑥 2 )𝑑𝑥 + 2 ∫1 𝑥 2 𝑠𝑒𝑐(𝑥 2 )𝑑𝑥 = 𝑡𝑎𝑛(1) = 1.5574 𝑅𝐻𝑆 = 𝜋 4 = 0.7853 From the above 𝐿𝐻𝑆 ≠ 𝑅𝐻𝑆 𝜋 tan(1) ≠ 4 Now LHS is not equal to the RHS hence the statement is false now we move to the next step for the result Step 3Result Result: from the above solution and analysis we conclude the given statement is false the LHS is not equal to the RHS and we mentioned the result below 0 0 𝜋 ∫1 𝑡𝑎𝑛(𝑥 2 )𝑑𝑥 + 2 ∫1 𝑥 2 𝑠𝑒𝑐(𝑥 2 )𝑑𝑥 ≠ false statement 4 so we calculate the true value is below 0 0 ∫1 𝑡𝑎𝑛(𝑥 2 )𝑑𝑥 + 2 ∫1 𝑥 2 𝑠𝑒𝑐(𝑥 2 )𝑑𝑥 = 𝑡𝑎𝑛(1) true statement Question #2 Does the function 𝑓(𝑥) = 1 𝑥 have an inverse function? Answer: Step 1 The horizontal line test only tells us the function is injective. Anyway, before we start talking about injectivity/surjectivity/biijectivity, we should always specify domains and target spaces. The function 𝑅 𝑓1 : → 𝑅 {0} 𝑓1 (𝑥) = 1 𝑥 is injective, but not surjective. The function 𝑅 𝑅 𝑓2 : → {0} 𝑓2 (𝑥) = {0} 1 𝑥 is bijective, and (𝑓2 )−1 = 𝑓2 The function 𝑓3 : (0, ∞) → 1 𝑓3 (𝑥) = 𝑥 is injective but not surjective. The function 𝑓_{4}: (0, ∞) → (0, ∞) 1 𝑓4 (𝑥) = 𝑥 is bijective, and (𝑓4 )−1 = 𝑓4 And so on. But note that typically, if we have a function 𝑓: 𝐴 → 𝐵 which is injective, it usually doesn't hurt to restrict it's target space to equal the image, and then that resulting function will be bijective. Question #3 Show that 1 𝑛 𝑛 ∑∞ 𝑗=1(1 − 𝑝𝑗 ) → 0 as 𝑛 → ∞ Answer: Step 1 From the inequality that OP observed, we get 𝑛 (1 − 𝑝𝑗 ) ≥ max{ 1 − 𝑛𝑝𝑗 , 0} and hence 𝑛 1 − (1 − 𝑝𝑗 ) ≤ min{ 𝑛𝑝𝑗 , 1} From this, we get 𝑛 1 ∞ 1 ∑𝑗=1(1 − (1 − 𝑝𝑗 ) ) ≤ ∑∞ 𝑗=1 min{ 𝑝𝑗 , } 𝑛 𝑛 Since the jth summand of the last sum is always bounded by 𝑝𝑗 and ∑∞ 𝑗=1 𝑝𝑗 converges, by the dominated convergence theorem   1 infty  1 lim  min{ p j , } =  = lim min{ p j , } =  min{ p j , 0} = 0 n → n → n n j =1 j =1 j =1 j =1 So by the squeezing theorem, 1 n ((1 − p j ) n ) = 0  n → n j =1 lim Question #4 Understanding a statement about the series 𝑆 = ∑∞ 2 1 𝑛 ln 𝑛 Answer: Step 1 By 𝑛 ln 𝑛 < 𝑛1+𝜖 , thus 1 1 ∑ > ∑ 1+𝜖 𝑛 ln 𝑛 𝑛 The series 1 ∑ 1+𝜖 𝑛 converges. This cannot be conclusive about the convergence of S, because S is bigger than something finite then we don't know if S is finite or infinite. Hence, you should use integral. Question #5 Prove that 𝑑𝑓(𝑐) − 𝑐𝑓(𝑑) 𝑓(𝜉) = ′ −𝜉 𝑓(𝑑) − 𝑓(𝑐) 𝑓 (𝜉) for some 𝜉 ∈ (𝑐, 𝑑) Answer: Step 1 According to Cauchy's version of Mean Value Theorem, there exists an 𝜉 ∈ (𝑐, 𝑑) , such that 𝑐 𝑐𝑓(𝑑)−𝑑𝑓(𝑐) 𝑓(𝑑)−𝑓(𝑐) = = 𝑑 𝑓(𝑑) 1 − 𝑓(𝑐) 𝑓(𝑑) = 𝑓(𝑐) 1 − 𝜉 ′ ) 𝑓(𝜉) 1 ′ ( ) 𝑓(𝜉) ( 𝑓(𝜉)−𝜉𝑓′ (𝜉) 𝑓2 (𝜉) 𝑓′ (𝜉) − 2 𝑓 (𝜉) =− 𝑓(𝜉)−𝜉𝑓 ′ (𝜉) =𝜉− 𝑓 ′(𝜉) 𝑓(𝜉) 𝑓 ′(𝜉) Question #6 I am trying to do excercise 4.1. in Continuous Time Markov Processes: An Introduction by Thomas Ligett. There I have to provide some counterexample of a function in 𝐶({0,1}𝑁 ) . However I don't know what these continuous functions are. The setting is the following: We have {0,1}𝑁 equipped with the product topology. According to the author we can metrize this space with the metric 𝜌(𝜂, 𝜁) = ∑ 𝛼(𝑥)|𝜂(𝑥) − 𝜁(𝑥)| 𝑥∈𝑁 with 𝛼(𝑥) is strictly positive and summable. So my question is: How do I prove that a function 𝑓: {0,1}𝑁 → 𝑅 is continuous with respect to the product topology. Can I just fix a sequence 𝛼 and then apply the 𝜖 − 𝛿 definition or do I have to account for something else? Answer: {0,1}𝑁 has the product topology. A sequence converges in this space iff each coordinate converges. Since the coordinates are just 0 or 1 this means each coordinate becomes constant after some stage. An example of dis-continuous function on this space is given by 𝑓((𝑥𝑛 )) = 𝑖𝑛 𝑓{𝑛: 𝑥𝑛 = 1} if 𝑥𝑛 = 1 for some n and 𝑓(0,0, , … ) = 0 . Note that 𝑓(𝑒𝑛 ) = 𝑛 if 𝑒𝑛 is the sequence qwith 1 in the n-th place and 0 elsewhere. Also, 𝑒𝑛 → (0,0, … ) . Clearly 𝑓(𝑒𝑛 ) does not tend to 𝑓(0,0, … ) so f is not continuous. Question #7 The proof that if f(x) is continuous at 𝑥 = 𝑐 , then 1/𝑓(𝑥) is continuous at 𝑥 = 𝑐 If f(x) is continuous at 𝑥 = 𝑐, 𝑓(𝑥) ≠ 0 then 1/𝑓(𝑥) is continuous at 𝑥 = 𝑐 . My proof: for any 𝜖 > 0 , we need to show there exists 𝛿 > 0 , such that whenever |𝑥 − 𝑐| < 𝛿, we have 1 − f ( x) 1 | − f ( x) | 1 | . f (c ) 1 f ( x ) − f (c ) |=| | f (c ) f ( x ) f (c ) 1 = |𝑓(𝑥) − 𝑓(𝑐)| |𝑓(𝑥)𝑓(𝑐)| 1 1 = |𝑓(𝑥) − 𝑓(𝑐)| |𝑓(𝑥)| |𝑓(𝑐)| Since f is continuous at c, thus |𝑓(𝑥) − 𝑓(𝑐)| < 𝜖. I need to find an upper bound of |𝑓(𝑥)| . Because |𝑓(𝑥) − 𝑓(𝑐)| < 𝜖, we then have |𝑓(𝑐)| = |𝑓(𝑐) − 𝑓(𝑥) + 𝑓(𝑥)| ≤ |𝑓(𝑐) − 𝑓(𝑥)| + |𝑓(𝑥)| . < 𝜖 + |𝑓(𝑥)| Thus, we find |𝑓(𝑥)| > |𝑓(𝑐)| −. Equivalently, could be negative, how to remedy?) Finally, (ignore not rigorous part) 1 |𝑓(𝑥)| 1 < |𝑓(𝑐)|−𝜖 . (Not rigorous here, since |f(c)|-\epsilon | 1 1 1 − | f ( x ) f (c ) | f (c ) | − | f (c ) | Let 𝜖 be arbitrarily small, i.e., 𝜖 → 0+ then |𝑓(𝑐)| − 𝜖 → 𝑓(𝑐) , then | 1 1 − | f ( x ) f ( c ) | f ( c ) |2 Hence, if at the beginning, let |𝑓(𝑥) − 𝑓(𝑐)| < 𝜖|𝑓(𝑐)|2 , then we can find that | 1 1 − | f ( x ) f (c ) The several last lines is not elegant. Maybe some small mistakes. How to remedy? Answer: You can start by taking 𝛿0 such that |𝑓(𝑐)| |𝑥 − 𝑐| < 𝛿0 ⇒ |𝑓(𝑥) − 𝑓(𝑐)| < . 2 𝑓(𝑐) Therefore |𝑥 − 𝑐| < 𝛿0 ⇒ |𝑓(𝑥)| > 2 1 1 | f ( x ) − f (c ) | − | 2 and so | x − c |  0 | f ( x ) f (c ) f 2 (c ) So, if you take 𝛿1 > 0 such that 𝜉𝑓 2 (𝑐) |𝑥 − 𝑐| < 𝛿1 ⇒ |𝑓(𝑥) − 𝑓(𝑐)| < , 2 then | x − c | min{ 0 , 1} | 1 1 − |  . f ( x ) f (c ) Question #8 Investigate the convergence or divergence of the sequence 𝑎1 = √𝑐, 𝑎2 = √𝑐 + 𝑎1 , 𝑎𝑛+1 = √𝑐 + 𝑎𝑛 , 𝑐 > 0. I proved it is increasing. But I could not prove it is bounded above or unbounded. I guess it is divergence sequence. How can I do this? Answer: Claim: 𝑎𝑛 is increasing and bounded above by 𝑀= 1+√1+4𝑐 2 . Also lim 𝑎𝑛 = 𝑀 𝑛→∞ Proof: By induction on n \geq 1 . 𝑎2 = √𝑐 + √𝑐 > √𝑐 = 𝑎1 . Assume 𝑎𝑛 > 𝑎𝑛−1 , we show 𝑎𝑛+1 > 𝑎𝑛 . But 𝑎𝑛+1 = √𝑐 + 𝑎𝑛 > √𝑐 + 𝑎𝑛−1 = 𝑎𝑛 by the inductive step. So by induction 𝑎𝑛+1 > 𝑎𝑛 , ∀𝑛 ≥ 1 . We show that 𝑎𝑛 < 𝑀, ∀𝑛 ≥ 1 by induction also. For 1 + √1 + 4𝑐 𝑛 = 1, 𝑎1 = √𝑐 < 𝑀 ⇔ √𝑐 < ⇔ 2√𝑐 − 1 < √1 + 4𝑐 2 1 If 𝑐 < , we're done since 𝐿𝐻𝑆 < 0 < 𝑅𝐻𝑆 . 4 1 Otherwise, 𝑐 > ⇒ 2√𝑐 − 1 > 0, and squaring both sides: 4𝑐 − 4√𝑐 + 1 < 1 + 4𝑐 ⇔ −4√𝑐 < 0 is clearly 4 true. Thus 𝑎1 < 𝑀 . Assume 𝑎𝑛 < 𝑀 , you have 𝑎𝑛+1 = √𝑐 + 𝑎𝑛 < 𝑀 ⇔ 𝑎𝑛 < 𝑀 2 − 𝑐 = 1+2√1+4𝑐+1+4𝑐 4 −𝑐 = 1+√1+4𝑐 2 =𝑀 which is true by inductive step. So 𝑎𝑛 < 𝑀, ∀𝑛 ≥ 1 . Thus the claim is verified. This follows that the limit exists and call it M. Then 𝑀 = √𝑐 + 𝑀 ⇒ 𝑀2 = 𝑐 + 𝑀 ⇒ 𝑀2 − 𝑀 − 𝑐 = 0 ⇒ 𝑀 = 1±√1+4𝑐 2 Since 𝑎𝑛 > 0, ∀𝑛 ≥ 1 ⇒ 𝑀 ≥ 0 ⇒ 𝑀 = 1+√1+4𝑐 2 Question #9 Proving that 𝑥 − 𝑥 2 is a contraction without mean value theorem Answer: Sure, it's fairly straightforward. Consider that 𝑓(𝑥) − 𝑓(𝑦) = (𝑥 − 𝑥 2 ) − (𝑦 − 𝑦 2 ) = (𝑥 − 𝑦). −(𝑥 2 − 𝑦 2 ) = (𝑥 − 𝑦)(1 − 𝑥 − 𝑦) Then in particular, we have 𝑑(𝑓(𝑥), 𝑓(𝑦)) = |𝑥 − 𝑦||1 − 𝑥 − 𝑦| = 𝑑(𝑥, 𝑦)|1 − 𝑥. −𝑦| Now we see that −1 ≤ 1 − 𝑥 − 𝑦 ≤ 1 . In particular, the extremes are only achieved when 𝑥 = 𝑦 ∈ {0,1} , so when 𝑥 ≠ 𝑦 , we have −1 < 1 − 𝑥 − 𝑦 < 1 . In this case, we see that |1 − 𝑥 − 𝑦| < 1 . Therefore, we see that when 𝑥 ≠ 𝑦 , we have 𝑑(𝑓(𝑥), 𝑓(𝑦)) = 𝑑(𝑥, 𝑦)|1 − 𝑥 − 𝑦| < 𝑑(𝑥, 𝑦). Now suppose the map were a strict contraction. Then take some 𝑐 ∈ (0,1) such that for all 𝑥 ≠ 𝑦 , we have 𝑑(𝑓(𝑥), 𝑓(𝑦)) < 𝑐𝑑(𝑥, 𝑦) Then let 𝑘 = 1+𝑐 2 ; then 0 < 𝑐 < 𝑘 < 1 . And we see that 𝑑(𝑓(1), 𝑓(𝑘)) = 𝑑(1, 𝑘)|1 − 1 − 𝑘| = 𝑑(1, 𝑘)|𝑘| . This is = 𝑑(1, 𝑘)𝑘 > 𝑑(1, 𝑘)𝑐 > 𝑑(𝑓(1), 𝑓(𝑘)) a contradiction. Question #10 If 𝑓𝑛 ≥ 0 a.e. and 𝑓𝑛 → 𝑓 in measure then 𝑓 ≥ 0 a.e. Answer: {𝑥: 𝑓(𝑥) < 0} ⊆∪𝑘 {𝑥: 𝑓𝑛𝑘 (𝑥) < 0} ∪ 𝐴 where A is the set of points where (𝑓𝑛𝑘 (𝑥)) does not converge to 𝑓(𝑥) > Question #11 How can I find the order of the pole 𝑧 = 𝜋 2 for 𝑓(𝑧) = (2 1 log(𝑧))(1−sin(𝑧)) ? I know the answer should be 2, but I can't solve it, mostly due to poor understanding of the pole order theory. As far as I've understood, the order should be equal to the index of the first non-zero member of the expansion in a series. However, I don't know how to expand this function in the singularity for a Taylor series, and a geometrical series doesn't seem to do the trick Answer: First, note that 2 log(𝑧) does not vanish at 𝑧 = So consider simply 1 1−sin(𝑧) 𝜋 𝜋 2 , so we can ignore it; it does not contribute to the pole. 𝜋 . We want to write things in terms of 𝑤 = 𝑧 − , 2 so let sin(𝑧) = sin (𝑤 + ) = cos(𝑤) , where at the end I have used a simple trig formula. Then we can expand cos(𝑤) = 1 − 1 1−cos(𝑤) = 𝑤2 2 1 𝑤4 +⋯ 24 − = 2 𝑤2 2 + 1 𝑤4 24 − ⋯, so 1 𝑤2 +⋯ ) 2 24 𝑤2 ( − Thus, we can see that the order of the pole is 2. Question #12 How do I proof that this is a metric in R? I've been struggling to proof that: | x y − |= 0  x = y . This way ⇐ has been easy but i don't know how to proof the other way. 1+ | x | 1+ | y | I tried a distinction of cases but it didn't really work out the way I wanted. Has anyone got a tip for me? Answer: One observation is that if | x y − |= 0 then 𝑥, 𝑦 has to be both positive or both negative. Thus 1+ | x | 1+ | y | we can consider two different cases. Case 1: Both positive. Thus we have | x y − |= 0 . In other words, 1+ x 1+ y derivatives if necesary, because 𝑓(𝑥) = 𝑥 1+𝑥 𝑥 1+𝑥 = 𝑦 1+𝑦 . Now use is one to one, we have x=y . Alternatively, suggested by Dan, one could just multiply through to have 𝑥 + 𝑥𝑦 = 𝑦 + 𝑦𝑥 and thus 𝑥 = 𝑦 . (We are allowed to multiply through since neither x nor y were assumed to be negative. Case 2: Both negative. This is similar to Case 1. Question #13 I try to study the convergence or divergence of the series ∑∞ 𝑛=2 of (−1)𝑛 . We can see that this series diverges absolutely, because ∑∞ 𝑛=2 𝑛+(−1)𝑛 1 ∑𝑛=2 which 𝑛 (−1)𝑛 ∑∞ 𝑛=2 𝑛+(−1)𝑛 of 1 |𝑛+(−1)𝑛| is a rearrangement equals to +∞ . Consequently, we can't say for sure that the rearrangement (−1) ∑∞ 𝑛=2 𝑛 𝑛 converges. That's the part where I am stuck. How exactly can I study the convergence (or not) of the first series? Should I apply Cauchy's criterion and prove that 𝑠𝑛 = ∑𝑛𝑘=2 (−1)𝑘 𝑘+(−1)𝑘 , 𝑛 ∈ 𝑁 is a basic sequence? Answer: Without giving a formal proof of it (or even a formal statement of the theorem), the following is true: You can do any bounded amount of rearranging of an infinite series without affecting its (conditional) convergence/divergence. For the series at hand, note that ∞ (−1)𝑛 −1 1 −1 1 −1 1 ∑ = + + + + + +⋯ 𝑛 𝑛 + (−1) 3 2 5 4 7 6 𝑛=2 1 1 1 1 1 1 2 3 4 5 6 7 = − + − + − + ⋯. where the ''bounded'' rearranging is to swap pairs of adjacent terms. The rearranged sequence now meets all the criteria of Leibniz's test, so the series is conditionally convergent. (The rearrangement makes it easy to see that the sum is 1 − ln 2 .) A somewhat more formal definition of ''bounded'' rearrangement is that no term is the rearranged series is ever more than B positions from where it appears in the original series, for some fixed positive number B. (In this example, 𝐵 = 1 .) Somewhat more formally, a rearrangement ∑ 𝑎𝜋(𝑛) or ∑ 𝑎𝑛 , where 𝜋: 𝑁 → 𝑁 is a bijection (aka permutation), is ''bounded'' (by B) if |𝜋(𝑛) − 𝑛| ≤ 𝐵 for all n. Question #14 Let 𝑓: (𝑎, 𝑏) → 𝑅 be a differentiable map and let 𝑥0 ∈ (𝑎, 𝑏) be a critical point. Prove that if there exists 𝛿 > 0 such that 𝑓 ′ (𝑥) ≥ 0 for all 𝑥 ∈ (𝑥0 − 𝛿, 𝑥0 ) and 𝑓 ′ (𝑥) ≤ 0 for all 𝑥 ∈ (𝑥0 , 𝑥0 + 𝛿) then 𝑥0 is a local maximum of f. Answer: Step 1 Suppose 𝑥0 is not a local maximum. Then we should be able to find x such that |𝑥 − 𝑥0 | < 𝛿 and 𝑓(𝑥) > 𝑓(𝑥0 ) . Now apply the mean value theorem to relate the difference 𝑓(𝑥) − 𝑓(𝑥0 ) to the derivative of f evaluated at some point near 𝑥0 , and use what you know about the sign of the derivative near 𝑥0 . Question #15 Find the derivative of a function with new variable Suppose 𝑓(𝑥 − 3) = (𝑥 − 2)2 and find the 𝑓 ′ (𝑥 2 + 5) Answer: Step 1 You calculated the derivative of the composite function 𝑥 ↦ 𝑓(𝑥 2 + 5) , and evaluated it at x, rather than calculating the derivative of the function 𝑡 ↦ 𝑓(𝑡) , and evaluating it at 𝑥 2 + 5 . Note that by definition ′ (𝑥 ↦ 𝑓(𝑥 2 + 5)) (𝑥) = lim 𝑓((𝑐+ℎ)2 +5)−𝑓(𝑥 2 +5) ℎ ℎ→0 whereas ′ (𝑡 ↦ 𝑓(𝑡)) (𝑥 2 + 5) = lim = lim 𝑓(𝑥 2 +5+ℎ)−𝑓(𝑥 2 +5) 𝑓(𝑡+ℎ)−𝑓(𝑡) ℎ ℎ→0 |𝑡=𝑥 2+5 ℎ ℎ→0 Question #16 Calculate the sum of the series ∞ ∑ 𝑛=1 𝑛 + 12 𝑛3 + 5𝑛2 + 6𝑛 Answer: Step 1 As you said, for every 𝑁 ∈ 𝑁 ⋆ , one has 2 5 3 𝑁 ∑𝑁 + 𝑛=1 𝑎𝑛 = ∑𝑛=1 − 𝑛 1 𝑛+2 1 𝑛+3 𝑁+2 𝑁+3 = 2 ∑𝑁 𝑛=1 − 5 ∑𝑛=3 + 3 ∑𝑛=4 𝑛 1 𝑛 1 1 𝑛 1 = 2 (1 + + + ∑𝑁 𝑛=4 ) 1 −5 ( + 3 𝑁 2 3 1 ∑𝑁 𝑛=4 𝑛 𝑛 + 1 𝑁+1 + 1 𝑁+2 ) 1 1 1 1 + + + ) 𝑛 𝑁+1 𝑁+2 𝑁+3 𝑛=4 1 1 = 2 − 5( + ) 𝑁+1 𝑁+2 +3 (∑ +3 ( 1 𝑁+1 + 1 𝑁+2 + 1 𝑁+3 ) Now just let N tend to +∞ to see that ∞ ∑ 𝑎𝑛 = 2 𝑛=1 Question #17 Verification of the limit of a succession ln(𝑛 + 1) =0 𝑛→+∞ 2𝑛 lim Procedure. | ln(n + 1) | 2n Answer: Step 1 Let (𝑎𝑛 )𝑛≥1 be a sequence defined by 𝑎𝑛 = ln(𝑛+1) 𝑐𝑛 𝑐 ∈ 𝑅+× By definition, 𝑎𝑛 ∼𝑛→+∞l ⇔ ∀𝜖 > 0∃𝑁 ∈ 𝑁, 𝑛 ≥ 𝑁: |𝑎𝑛 − l| < 𝜖 that is, ln(𝑛 + 1) ∼𝑛→+∞ 0 ⇔ ∀𝜖 > 0∃𝑁 ∈ 𝑁, 𝑛 ≥ 𝑁: 𝑐𝑛 ln(𝑛+1) | − 0| < 𝜖 𝑐𝑛 Since that ln(𝑛 + 1) ln(𝑛 + 1) 𝑛+1 𝑛+1 | − 0| = | |≤| |= 𝑐𝑛 𝑐𝑛 𝑐𝑛 𝑐𝑛 1 <𝜖⇔𝑛> 1 𝑐 (𝜖 − ) 𝑐 Therefore, 1 × Let 𝜖 > 0 and 𝑁 ∈ 𝑁 such that 𝑛 > 1 with 𝑐 ∈ 𝑅+ . Then for all 𝑛 ∈ 𝑁 such that 𝑛 ≥ 𝑁 , we have 𝑐(𝜖− ) 𝑐 ln(𝑛 + 1) ln(𝑛 + 1) 𝑛+1 𝑛+1 | − 0| = | |≤| |= 𝑐𝑛 𝑐𝑛 𝑐𝑛 𝑐𝑛 1 1 1 1 = + < +𝜖− =𝜖 𝑐 𝑐𝑛 𝑐 𝑐 Note: We know that for all 𝑥 ∈ 𝑅+× we have that 𝑥−1 ≤ ln(𝑥) ≤ 𝑥 − 1 𝑥 Reference: For all 𝑥 ∈ 𝑅+× we have Question #18 Calculate the improper integral ∞ ∫ 0 1 cos 𝜃 𝑒 sin(sin 𝜃) 𝑑𝜃 𝜃 Answer: 𝑥−1 𝑥 ≤ ln(𝑥) ≤ 𝑥 − 1 Step 1 Define the sine integral by 𝑥 sin 𝑡 𝑆𝑖(𝑥) = ∫ 𝑑𝑡 𝑡 0 Using integartion by parts, it can be proved that 𝜋 1 2 𝑥 𝑆𝑖(𝑥) = + {𝑂} ( ) as 𝑥 → ∞ Now note that ∞ 𝑒 cos 𝜃 sin sin 𝜃 = 𝐼𝑚 (𝑒 𝑒 1𝜃 − 1) = ∑ 𝑛=1 1 sin(𝑛𝜃) 𝑛! Then by the Fubini's theorem, for 𝑅 > 0 𝑅 𝑒 cos 𝜃 sin sin 𝜃 ∫0 ∞ =∑ 𝑛=1 ∞ =∑ 𝑛=1 ∞ 𝜃 𝑅1 𝑑𝜃 = ∫0 𝜃 ∑∞ 𝑛=1 sin(𝑛𝜃) 𝑛! 𝑑𝜃 1 𝑅 sin(𝑛𝜃) ∫ 𝑑𝜃 𝑛! 0 𝜃 1 𝑆𝑖(𝑛𝑅) 𝑛! 1 𝜋 1 ( + {𝑂} ( )) 𝑛! 2 𝑛𝑅 𝑛=1 𝜋 1 = (𝑒 − 1) + {𝑂} ( ) 2 𝑅 So by letting 𝑅 → ∞, the integral converges to ∞ cos 𝜃 𝑒 sin sin 𝜃 𝜋 ∫ 𝑑𝜃 = (𝑒 − 1) 𝜃 2 0 =∑ Question #19 How do I prove that 𝑓(𝑥) = 𝑥 2 − 4𝑥 + 3 is injective at (−∞, 2) and (2, ∞) Answer: Step 1 If 𝑓(𝑥) = 𝑓(𝑦) then 𝑥 2 − 4𝑥 = 𝑦 2 − 4𝑦. Hence 𝑥 2 − 𝑦 2 − 4(𝑥 − 𝑦) = 0, which means (𝑥 − 𝑦)(𝑥 + 𝑦 − 4) = 0. This is true if and only if 𝑥 = 𝑦 or 𝑥 + 𝑦 − 4 = 0 . Now if 𝑥, 𝑦 > 2 , then 𝑥 + 𝑦 > 4 , so the only possibility is 𝑥 = 𝑦 . Same for 𝑥, 𝑦 < 2 (try). Question #20 Function f with |𝑓| is Lebesgue integrable but f isn't locally Lebesgue integrable Answer: Step 1 In point of fact a function 𝑓: 𝑅 → 𝑅 such that |𝑓| is integrable will be integrable (locally and/or otherwise) if and only if it is measurable. Your f is not measurable because 1 𝐴 = 𝑓 − 1 [ , ∞), 2 and therefore it isn't integrable.