Ph.D. Katarína Bellová Page 1 Mathematics 1 (10-PHY-BIPMA1) EXAM SOLUTIONS, 20 February 2018

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1 Ph.D. Katarína Bellová Page Mathematics 0-PHY-BIPMA) EXAM SOLUTIONS, 0 February 08 Problem [4 points]: For which positive integers n does the following inequality hold? n! 3 n Solution: Trying first few positive integers, we get the following table: n n! n We see that for n = 5, n! > 3 n. If we continue the table further, for each following integer n +, we multiply n! by n + whereas we multiply 3 n by 3, which is less than n + as n 5). Hence, the first line will stay larger than the second line. Formally, we can prove this by induction: We claim that for any n 5, n! > 3 n. As the first induction step, for n = 5, n! = 0 > 8 = 3 n holds true. For the second step, assume that n! > 3 n holds for some n = 5 and let us prove it for n = +. Using the induction assumption and + > 3, we have + )! =! + ) > 3 + ) > 3 3 = 3. Hence, n! > 3 n is proved also for n = +, and thus for any n 5. To conclude, we need to decide the validity of our statement for the first few n s for which the induction does not wor!) based on the table. We then see that n! 3 n holds for n = and any n 5. Problem [6 points]: Let a =, a n+ = a n + for n. Prove that the sequence a n ) n is a) bounded; b) convergent. Solution: a) By computing the first few sequence terms, we can guess that the sequence stays bounded between and. Formally, this is best proved by indution: for n =, a = [, ]. Let us now assume a n [, ] for some n = and prove it for n = +. By the recurrent definition of a + and induction assumption, we have a + = a + + >, a + = a + + =. Hence, a n [, ] holds also for n = +, and thus for any n.

2 Ph.D. Katarína Bellová Page Mathematics 0-PHY-BIPMA) b) Recall that any bounded monotone sequence is convergent. In light of a), it is thus sufficient to prove that a n is monotone. Indeed, a n is monotone increasing: this is consecutively equivalent to a n+ a n a n + a n a n a n. As the last inequality is true for any n by part a), it is also true that a n+ a n for any n, and the sequence a n ) is monotone increasing and thus convergent. Knowing that the sequence is convergent, it is now also easy to compute the it which was not part of the problem). Let n a n = L. Then also n a n+ = L and passing to the it in the recurrence formula a n+ = an + gives which leads to L =. Problem 3 [4 points]: Compute the it L = L +, Solution: Using the nown its e e cos) = e cos). e = 4 e = 4 = and cos =. cos) ) 4 ) cos) ) =, we have We used several algebraic rules, which were justified since the last line maes sence, as well as substitution as 0, 0 and 0). Alternatively, the problem can be solved by l Hospital rule. Problem 4 [6 points]: Is the following function f continuous on R? Is it differentiable on R? { ln for 0, f) = 0 for = 0. Solution: First, function f is clearly continuous and differentiable for any 0. Let us analyse the properties in 0 = 0. We have f) = ln = 0 = f0),

3 Ph.D. Katarína Bellová Page 3 Mathematics 0-PHY-BIPMA) so f is continuous in 0; if one does not remember the well-nown it + ln = 0 and thus ln = + ln = 0), it can be easily derived by l Hospital rule: ln ln = + + = + Thus, f is continuous on R. As for differentiability on 0, we proceed from definition: f 0) = f) f0) 0 = ln = + = 0. = ln =. Hence, f is not differentiable in = 0, and thus not differentiable on R. Problem 5 [3 points]: Give the definition domain of the function f) = cose ) and compute its first and second derivative. Solution: As both functions e and cos are defined on R, so is their composition and the definition domain of f is R. We compute the derivatives by chain rule and product rule - again, they eist on the whole R: f ) = cose )) = sine )e, f ) = sine )e ) = cose )e e sine )e = cose )e sine )e. Problem 6 [4 points]: Prove that the function f) = sin + 4 attains a global minimum on R. Solution: First of all, f is continuous and we now that continuous functions attain their minima on any closed, bounded interval. We thus want to reduce the definition domain R onto a bounded interval I, and show that outside of I, function f will be larger than or equal to the minimum on I - then the minimum of f on I will also be a global minimum of f on R. Notice that f0) = 0, while for large the term 4 maes the function f large, and sin cannot change it that much. Indeed, as sin for any R, we get that = f) = sin = 0 = f0). Thus, we see that outside of I = [, ], the function will never be smaller than the value at 0, in particular the minimum of f on I which must be smaller or equal than f0), as 0 I) must be equal to the global minimum of f on R. Since I is bounded and closed and f is continuous on I, we now that the minimum of f on I is attained. Problem 7 [4 points]: Compute the Taylor polynomial of third order of the function f) = e sin at 0 = 0. Solution: Using the nown Taylor epansions for e and sin, we have e sin = + + ) + o ) ) o 4 ) = o 3 ) o 3 ) = o 3 ).

4 Ph.D. Katarína Bellová Page 4 Mathematics 0-PHY-BIPMA) Hence, P 3 ) = Alternatively, one can compute f0), f 0), f 0) and f 0) and plug them into P 3 ) = f0) + f 0) +! f 0) + 3! f 0) 3. Problem 8 [4 points]: Compute the indefinite integral sin d. Solution: Integrating by parts twice, we get sin d = cos + cos = cos + sin sin = cos + sin + cos + C. Problem 9 [4 points]: Compute the improper integral ) d. Solution: Using the substitution y = +, dy = d, we get ) d = dy y = + y = 0 ) =. Problem 0 [4 points]: For which a R does the following series converge? = ln + )) a Solution: First notice that ln + ) > 0 for any, so the general power is welldefined. Since =, we have ln+) )) ln + a ln )) a + a = = ) a =. con- Hence, by comparison criterion, = ln + verges, which is if and only if a >. Problem [4 points]: Epress + i) 08 in algebraic form. )) a converges if and only if = a

5 Ph.D. Katarína Bellová Page 5 Mathematics 0-PHY-BIPMA) Solution: For computing powers, the eponential/trigonometric form is the most convenient. As + i) = and arg + i) = π, we have 4 + i) 08 = e i π 4 ) 08 = ) 08 e i π 4 08 = 009 e πi 5+i π = 009 i. This is the algebraic form we were looing for real part vanishes). Alternatively, one can use the trigonometric form + i) = cos π 4 + i sin π 4 ) and compute the power by de Moivre s formula. Problem [6 points]: a) For what values of a R do the following vectors in R 3 form a basis? e =, 0, 0), e =,, ), e 3 = 0,, a) b) For any such a, find the coordinates of the vector v = 0, 0, ) in the above basis. Solution: a) Three vectors in R 3 form a basis if and only if they are linearly independent. As vectors e and e are not collinear one is not a scalar multiple of the other), they are linearly independent and for linear independence of e, e, e 3, we only need to assure that e 3 is not a linear combination of e, e. Let us find out when e 3 is a linear combination of e, e : then there must eist α, β R such that 0,, a) = α, 0, 0) + β,, ); in other words, 0 = α+β, = 0+β, a = 0+β. The first two equations give β =, α =, and the third gives a =. Hence, vectors e, e, e 3 form a basis if and only if a. b) For a, we need to epress v as a linear combination of e, e, e 3 : This leads to 0, 0, ) = α, 0, 0) + β,, ) + γ0,, a). 0 = α + β 0 = β + γ = β + aγ. From the first two equations, α = β = γ. Plugging this into the third equation gives = β a), i. e. β = notice that a!). The coordinates of v in the basis e a, e, e 3 are then α, β, γ) = a, a, a Problem 3 [4 points]: Let T : R 3 R 3 be given as T α, β, γ)) = α, α, β). Find the dimensions of the ernel Ker T ) and of the image Im T ). Solution: First notice that T is a linear map. The ernel of T consist of all vectors v = α, β, γ) such that T v) = 0, i. e. α, α, β) = 0, 0, 0). This happens for all v s such that α = β = 0, and γ can be arbitrary. Epressing this as vector space, Ker T = {γ0, 0, ) : γ R}. We see that the single vector 0, 0, ) forms a basis of Ker T, and thus dimker T ) =. As dimker T ) + dimim T ) = dim R 3 = 3, we must have dimim T ) =. Alternatively, it is also easy to find a basis of Im T, e. g. as,, 0), 0, 0, ).) ).