Math 8A: Homework Solutios Due: Jue. Determie the limits of the followig sequeces as. a) a = +. lim a + = lim =. b) a = + ). c) a = si4 +6) +. lim a = lim = lim + ) [ + ) ] = [ e ] = e 6. Observe that si 4 + 6) for all so that lim which demostrates that lim a = 0. ) d) a = l. + + Observe that + ) a = l + so that It follows that + a + + lim a lim + 0 lim a 0 = l + ) l + ) < l4) l + ) lim a < l 4) lim l + ) =. lim a =.
. Show that the followig equatios have at least oe solutio i the give itervals. a) x six) + x ) = 0 i [, π]. Let fx) = x six) + x ). Observe the that f is cotiuous o [, π] ad that f) = 4 si4) < 0 as π < 4 < π) while fπ) = π siπ) + π ) > 0. As f) < 0 < fπ), it follows from the itermediate value theorem that there exists some c, π) such that fc) = 0. b) e x sec πx) = i [, ]. Observe that we caot apply the IVT directly to a fuctio ivolvig sec ) as it is ot cotiuous at some poits i the iterval. We therefore rewrite the equatio as e x cos πx) = e x cos πx) = 0. Let fx) = e x cos πx). Observe the that f is cotiuous o [, ] ad that f ) = e cos π) = e > 0 while f) = e cos π) = e < 0. As f) < 0 < f ), it follows from the itermediate value theorem that there exists some c, ) such that fc) = 0.. Defie f : R R by fx) = { x si x) x 0 0 x = 0 a) Show that f is cotiuous at x = 0. Observe that f is defied at x = 0. We ext eed to show that lim x 0 fx) exists. Observe that for x 0, si ) x x x si x ) x lim x 0 x lim x 0 x si x) lim x x 0 0 lim x 0 x si x) 0 so that lim x 0 fx) exists ad equals zero. Sice lim x 0 fx) = 0 = f0), we coclude that f is cotiuous at x = 0. b) Is f differetiable as well at x = 0? Note that fx) f0) lim x 0 x 0 x si/x) 0 = lim x 0 x ) = lim x si. x 0 x By a argumet similar to the oe used i a), this limit ca also be show to exist ad equal zero. We coclude that f is ideed differetiable at x = 0 ad that f 0) = 0.
4. Prove that a) if x >, the lx) < x. Fix x > ad observe that fy) = ly) is cotiuous o [, x] ad differetiable o, x). The mea value theorem the states that at some c, x), f c) = fx) f) x c = lx) x. As c >, we get c < so lx) x < lx) < x. b) the graph of fx) = x 5 + 5x + c crosses the x-axis exactly oce, regardless of the value of the costat c. Observe first that ad f c ) = c ) 5 + 5 c ) + c = c + ) 5 + c 5 c ) 5 < 0 f c + ) = c + ) 5 + 5 c + ) + c = c + ) 5 + c + 5 c ) + 5 > 0 while f is cotiuous o [ c, c + ]. The itermediate value theorem the implies that there is at least oe z c, c + ) such that fz) = 0. I order to show that f crosses the x-axis o more tha oce, we assume the cotrary, i.e., suppose there are a, b R such that fa) = fb) = 0. Sice f is cotiuous o [a, b] ad differetiable o a, b), Rolle s theorem implies that there is some y a, b) such that f y) = 0. Note however that f x) = 5x 4 + 5 so f y) = 5y 4 +5 = 0 which however has o solutio. The cotradictio proves that there is exactly oe poit at which the graph of f crosses the x-axis. 5. Fid the Taylor polyomial of degree 5 about the poit x = 0 for the followig fuctios: a) fx) = e x. Observe that f x) = xe x, f x) = 4x + )e x, f ) x) = x4x + ) + 8x)e x, f 4) x) = x8x + x) + 4x + )e x ad f 5) x) = x6x 4 + 48x + ) + 64x + 96x)e x so that f0) =, f 0) = 0, f 0) =, f ) 0) = 0, f 4)0) = ad f 5) 0) = 0 so that P 5 x) = + x + x4 is the required polyomial. A much simpler way of gettig the same aswer is by recallig that the Taylor series of e x about x = 0 is + x + x / + x /6 +..., replacig x by x ad elimiatig all terms higher tha x 5.
b) fx) = l + x). Observe that f x) = + x), f x) = + x), f ) x) = + x), f 4) x) = 6 + x) 4 ad f 5) x) = 4 + x) 5 so that f0) = 0, f 0) =, f 0) =, f ) 0) =, f 4) 0) = 6, f 5) = 4 so the required polyomial is P 5 x) = x x + x 4 x4 + 5 x5. c) fx) = cos x). Observe that f x) = cosx) six) = six), f x) = cosx), f ) x) = 4 six), f 4) x) = 8 cosx) ad f 5) x) = 6 six) so that f0) =, f 0) = 0, f 0) =, f ) 0) = 0, f 4) 0) = 8, f 5) = 0 so the required polyomial is P 5 x) = x + x4. Aother way we ca get this is by squarig the Taylor series for cosx) about x = 0 ad droppig terms higher that x 5. Yet aother way is by usig the idetity cos x) = +cosx) ad replacig x by x i the Taylor series for cosx). 6. Fid the third Taylor polyomial P x) for the fuctio fx) = x + about x 0 = 0. Approximate 0.7 ad., fid upper bouds for the errors ad compare with actual errors. Observe that f x) = x + ) /, f x) = 4 x + ) /, f ) x) = 8 x + ) 5/, f 4) x) = 5 6 x + ) 7/ so that f0) =, f 0) =, f 0) = 4, f ) 0) = 8. We therefore have P x) = + x 8 x + 6 x. Note the that 0.7 = f 0.) P 0.) =.04885. The correspodig remaider term is P 0.) = 0.87065 while. = f0.) R x) = f 4) c) x 4 = 5 4! 8 + cx)) 7/ x 4 4
where cx) is some umber betwee 0 ad x. For x = 0., we have 0. < cx) < 0 so R 0.) < 5 8 0.)4 0.) 7/ =.0559 0. Likewise, for x = 0., we have 0 < cx) < 0. so R 0.) < 5 8 0.)4 + 0) 7/ =.9065 0 6 Note that the actual errors are f 0.) P 0.) = 4.0475 0 4 P 0.) =.6580 0 6. 7. Let fx) = x ) lx) ad x 0 =. a) Fid the third Taylor polyomial P for f about x 0. Note that f x) = lx) +, f x) = +, f )x) = x x x x x f) = 0, f ) = 0, f ) =, f ) ) = so that P x) = x ) x ). ad f0.) so that b) Use P to approximate f). Fid a upper boud for the error i the approximatio ad compare it to the actual error. Sice f 4) x) = + 6, we have x x 4 R ) = 4! f) P ) = 0.5. c)) + 6 c)) 4 ) ) 4 for some c) where < c) <. It follows that R ) < )4 4 ) + 6 ) = 0.97. ) 4 The actual error meawhile is f) P ) = 0.0407. c) Approximate fx) dx usig P i place of f. fx) dx = x ) x ) dx [ x ) x )4 8 = ) =. ].5 5
d) Fid a upper boud for the error i c). We require a upper boud o ɛ = fx) P x) dx. Note that ɛ = fx) P x) dx fx) P x) dx. ) Recall that for every x [,.5], there exists some cx) betwee x ad such that ) fx) P x) = R x) = 4! cx)) + 6 x ) 4. cx)) 4 For x [, ), we have x < cx) < so that x)4 R x) < 4 ) + 6 ) = 4 ) 4 x)4 while for x,.5], we have < cx) < x.5 so that x)4 R x) < 4 ) + 6 ) = ) 4 x)4 Pluggig these i ) yields ɛ < 4 = 4 5 R x) dx + x) 4 dx + [ ] x) 5 5 R x) dx x) 4 dx [ x) 5 ].5 = 4 5 )5 5 )5 = 5 )5 = 0.0708. 8. BONUS: Suppose a rig is heated so that the temperature varies cotiuously over it. Show that there exists a pair of diametrically opposite poits also called atipodal poits) that have the same temperature. Parameterize the rig by θ, i.e., each poit o the rig correspods to a value of θ [0, π). The temperature o the rig T θ) is the a cotiuous fuctio of θ; i additio, we have T 0) = T π) sice both θ = 0 ad θ = π correspod to the same poit. Defie fθ) = T θ) T θ +π) for over [0, π]. The fuctio f calculates the temperature differece betwee all pairs of atipodal poits. We wish to show that this differece is zero for some pair, i.e., fc) = 0 for some c i [0, π]. Note that f must be ecessarily cotiuous o [0, π] sice T is cotiuous. Furthermore, f0) = T 0) T θ) ad fπ) = T π) T π) = T π) T 0) so that f0) = fπ). Note that if f0) = 0, we are doe. Otherwise, f0) is either positive or egative. We 6
ca assume without loss of geerality that it s the latter; the fπ) must be positive so we have f0) < 0 < fπ). The itermediate value theorem the guaratees that at some c 0, π), we must have fc) = 0 T c) = T c + π). 7