8.4 Problem Set 9 Solutions Total: 4 points Problem : Integrate (a) (b) d. ( 4 + 4)( 4 + 5) d 4. Solution (4 points) (a) We use the method of partial fractions to write A B (C + D) = + +. ( ) ( 4 + 5) ( ) 4 + 5 Making a common denominator yields the epression A( )( 4 + 5) + B( 4 + 5) + (C + D)( ) =. Plugging in =, we get B =. Taking the derivative and plugging in = yields A =. Solving for C and D, we have ( 4 + 5) + (C + D)( ) =. Comparing 3 terms, we see C =. Comparing terms, we see D =. Thus, we have d d d ( 4 + 4)( 4 + 5) = ( ) 4 + 5. (Give yourself a pat on the back if you noticed from the start that the terms 4+5 and 4 4 differed by and got this decomposition without solving for A,B,C, and D). Now the first integral is ( ) + C. To do the second integral, we complete the square and substitute u = d d d 4 + 5 = ( ) + = = arctan u + C = arctan ( ) + C. u + Thus, our final answer is d ( 4 + 4)( 4 + 5) = ( ) arctan ( ) + C.
(b) Hopefully, after doing part (a), everyone observed ( ) = and =. + If not, one can figure this out using partial fractions. Now, we have ( ) d 4 = 4 + d ( ) = log log + + + C. 4 t Problem : Let A = e dt. Epress the values of the following integrals in t+ terms of A: a t e t te t e (a) dt (b) dt (c) dt (d) e t log( + t)dt. a t a t + (t + ) Solution (4 points) For part (a), we substitute t = u + a to get e u a e u u du = e a u + du = e a A. For part (b), we substitute u = t to get eu For part (c), we integrate by parts to get For part (d), we integrate by parts to get du = A. u + t e e t e + (t + ) t + dt = + + A. t e t log( + t) e dt = e log() A. + t Problem 3: Let F () = f(t)dt. Determine a formula (or formulas) for computing F () for all real if f is defined as follows:
(a) f(t) = (t { + t ). } t if t (b) f(t) = t if t > (c) f(t) = e t. (d) f(t) = ma{, t }. Solution (4 points) (a) If t, then f(t) =. If t, then f(t) = 4t. Hence, if then F () =, and if then 4 F () = 4 t dt = 3. 3 (b) If, then t 3 t F () = f(t)dt = ( t )dt + ( + t)dt = t + t + 3 = + + + + = + +. 3 6 If, then t 3 3 F () = f(t)dt = ( t )dt = t =. 3 3 If, then t 3 t F () = f(t)dt = ( t )dt + ( t)dt = t + t 3 = + + = +. 3 6 (c) If, then F () = f(t)dt = e t dt = e t = e +. If, then F () = f(t)dt = e t dt = e t = e. (d) Note t if t and t if t. Hence, f(t) = if t and f(t) = t if t. Thus, for, we have 3 3 F () = f(t)dt = dt + t = + 3 + 3 = 3 3. 3
For, we have F () = f(t)dt = dt =. And for, we have 3 3 F () = f(t)dt = dt + t dt = + = +. 3 3 3 3 Problem 4: Use Taylor s formula to show (a) n ( ) k k sin() = + E n (), where E n () (k )! ( (b) cos() = k= n n + )!. n ( ) k k n+ + E n+ (), where E n+ () (k)! ( n + )!. k= Solution (4 points) (a) By Taylor s formula (Theorem 7.6), we know sin() = n (sin (l) l ()) l= l! + E n (). Note sin l () = ± sin() =, sin 4l+ () = cos() =, and sin 4l+3 () = cos() = k. Thus, our sum becomes n ( ) k k=. To approimate the error term, we (k )! use theorem 7.7 to conclude n+ E n () M ( n + )! where M = sup sin(n+) (c). c Since sin (n+) (c) = ± cos(c), we observe M, and the bound for our error term n+ becomes () as desired. E n (n+)! (b) This time we observe cos (l+) () = ± sin() =, cos (4l) () = cos() =, and cos (4l+) () = cos() =. Plugging this into Taylor s formula yields cos() = n ( ) k k k= (k)! + E n+ (). 4
To approimate the error term, we again use theorem 7.7 together with the bound cos (l) (c) for all l. This yields the estimate E n+ () n+. (n+)! Problem 5: Evaluate the following limits: (e + ) (e ) log( + ) (a) lim (b) lim. 3 cos() Solution (4 points) (a) We apply L Hopital s rule (Theorem 7.9). Observe that both the numerator and the denominator are differentiable and take the value zero d d 3 at. We observe d ((e + ) (e )) = e e + and d = 3. Each of these functions is differentiable and takes the value zero at. Differentiating again, we get d (e e + ) = e and d (3 ) = 6. Both of these functions are d d differentiable and take the value zero at. Finally, we differentiate both functions d d one more time to get d (e ) = e + e and d (6) = 6. Applying L Hoptal s rule three times, we get (e + ) (e ) e + e lim = lim =. 3 6 6 The last equality comes from observing that the numerator and denominator are continuous functions and plugging in =. (b) Again, we use L Hopital s rule (Theorem 7.9). Observe that both the numerator and denominator are differentiable at and take the value zero at. Computing, we get d (log( + ) ) = and d (+ ( cos()) = sin(). Both of these d ) d functions are differentiable at and take the value zero at. Computing again, we get d ( ) = and d sin() = cos(). Applying L Hopital s rule twice, d + (+) d we get lim log( + ) = lim /( + ) =. cos() cos() The last equality comes from observing that the numerator and denominator are both continuous at zero and plugging in =. Problem 6: (a) Compute the limit sin() e + 7 3 /6 lim sin () sin( 3 ) (b) Show that if <, then e + + / 3 /. 5
(c) Show that if t <, then the approimation t t 3 t 5 e d t + + 3 5 involves an error in absolute value no more than t 7 /4. Solution (4 points) For part (a), we apply L Hopital s rule four times. Let f() = sin() e + 7 3 /6 be the numerator and let g() = sin () sin( 3 ) be the denominator. Note f () = cos() e e + 7 /, f () = sin() 6e 4 3 e + 7, f () = cos() 6e 4 e 8 4 e + 7, f (4) () = sin() 6e + 3 (f ()), and f (5) () = cos() 6e + (f ()). One sees that f is 5 times differentiable at, f (k) () = for k =,,..., 4, and f (5) () = 59. Net, we write g() = h ()h () where h () = sin () and h () = sin( 3 ). We leave it as eercise to the reader to verify by induction that n n h (k) ()h ( n k) g (n) () = (). k k= Now, we compute h () = sin() cos(), h () = cos () sin (), h () = 3 cos( 3 ), h () = 6 cos( 3 ) 9 4 sin( 3 ), and h () = 6 cos( 3 ) 8 3 sin( 3 ) 36 3 sin( 3 ) 7 6 cos( 3 ). Note h () = h () = h () = h () = h () =. These facts together with the above formula prove that h (k) () = for k =,..., 4. Note h () =, h () = 6. Thus, we get g (5) =. Applying L Hopital s rule 5 times, we end up with sin() e + 7 3 /6 59 lim. sin = () sin( 3 ) c (b) By Taylor s theorem, we know e = + + 3 / + E 3 where E 3 = 6 e for some c between and. If, then c and e c e because e t is an increasing 3 function of t. Since e 3, we have E 3 as desired. (c) Plugging in in our formula from part (b), we get that e = + + + E 3 6 where E 3 whenever. Integrating, we get t t 3 t 5 e d t + + t7 3 4 whenever t. Note that we used the comparison theorem for integrals to bound t E 3 ()d. 6 4
Bonus: Give an eample of an infinitely differentiable function f such that f is not identically zero but T n f = for all n. Here T n f denotes the nth Taylor polynomial of f centered at zero. Solution (4 points) Consider the function { } e / if > f() =. if Clearly f is infinitely differentiable on the interval (, ). To show that f() is infinitely differentiable on the interval (, ), we show by induction on n that f is n times differentiable for > and f (n) () = p n ( )e / for some polynomial p n whenever >.. The base case n = follows from the definition of f. For the inductive step, we assume f (n) () = p n ( )e / and we show that there eists a polynomial p n+ such that f (n+) () = p n+ ( )e /. To do this, we simply differentiate f (n) () = p n ( )e /. By the chain rule and the product rule, f (n+) () = p n e / p n 3 e /. Setting p n+ (t) = t p (t) t 3 p n (t) completes the inductive step. n In particular, we have shown that f() is infinitely differentiable on the interval (, ). It remains to show that f is infinitely differentiable at. We need a lemma. Lemma: If p is a polynomial, then lim p e / =. Proof: Given ɛ >, we must find δ > such that < δ implies p e / < ɛ. n n Put t = /, and let p(t) = k= c kt k. Then p(t) (n + ) ma n k= { c k } t t m t if t. Note that e k k= by Taylor s formula (Theorem 7.6) and the k! bound on the error term (Theorem 7.7) together with the fact that the derivative of e t is never negative. If we choose m > n, then we see that e t t m and if (m)! t > (n + ) ɛ man k= { c k } =, we conclude that δ e t >. p(t) ɛ 7
Plugging in t = / and taking the inverse of the above equality, we note that p e / < ɛ whenever < δ. This proves the lemma. Back to showing that f (n) is differentiable at = for every n. Since f (n) () = p e / n, we note that f (n) () lim = lim qn e /. + + Here q n (t) = tp n (t). By the lemma, the above limit is zero. Since the limit from the right is obviously zero, we conclude that f (n) is differentiable at zero with derivative zero for all n. Finally, T n f = f (n) () n /n! by definition. Since f (n) = for all n, we have T n f = for all n. However, f is not identically zero. For instance, f() = /e. 8
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