STEP Support Programme. STEP III Hyperbolic Functions: Solutions

Transcription

1 STEP Spport Programme STEP III Hyperbolic Fnctions: Soltions Start by sing the sbstittion t cosh x. This gives: sinh x cosh a cosh x cosh a sinh x t sinh x dt t dt t + ln t ln t + ln cosh a ln ln cosh a cosh a + ln cosh a + cosh a + ln ln + + Usally yo cold have sed the reslt for the integral of a given in the formla book, For the next integral se t sinh x. This gives: x bt as this is a show that yo shold show how each stage is derived. cosh x sinh a + sinh x cosh x + t cosh x dt sinh a dt + t tan sinh a t tan sinh a For this second integral the answer is not given. The integral of formla book. a is given in the STEP + x For the Hence, first note that + sinh x + cosh x cosh x. We then have: cosh x sinh x + sinh x cosh x + sinh x cosh x + sinh x tan sinh a ln sinh x + sinh x sinh x cosh x cosh a cosh a + ln + STEP III Hyperbolics: Soltions

2 Then as a we have cosh a and sinh a. This means that and tan sinh a π. Hence we have: cosh x sinh x + sinh x π + ln as reqired. cosh a cosh a + For the last part, start by noting that cosh x ex + e x + and similarly sinh x. Using the sbstittion e x gives: cosh x sinh x + sinh x d + d 4 + d Hence from the previos reslt we have: 4 + d π ln. STEP III Hyperbolics: Soltions

3 Refer to the hints for the formlae sed which are in the formla book!. There are lots of different approaches, this is jst one possible method. Using integration by parts on T gives: artanh t t dt ln t artanh t + t ln t ln t ln + ln ln ln ln ln t t dt sing 6 ln t t dt sing ln + ln ln ln ln + ln ln ln t t dt V ln t ln t t ln t t dt dt t dt Comparing U with T and V, it wold be nice if I cold convert a limit of ln to one of. If ln implies t then it might be worth trying t e, which gives dt d e t. Using this sbstittion: ln ln ln sinh d ln V e d e ln t t t dt t ln t t dt ln t t dt The final thing we need to do is show that X is eqal to one of the other three. Looking at the limits t and x ln sggests that we might want to se a sbstittion of x ln t ln t or eqivalently t ex. STEP III Hyperbolics: Soltions

4 Starting with T we have: artanh t t dt ln ln ln artanh e x e x e x artanh e x ln ln + e x ln ln e x sing ln e x + e x ln ln e x e x mltiplying top and bottom of fraction by e x cosh x ln ln sinh x ln ln ln coth x ln X STEP III Hyperbolics: Soltions 4

5 Differentiating we have dy x x, bt this doesn t look immediately promising. We also have: y ln r ln r x x x x Since we have coth θ x, x coth θ cosech θ. Hence: We therefore have: Now we differentiate again: Since x coth θ, we have x x coth θ cosech θ cosh θ sinh θ sinh θ cosh θ dy dy d y d r cosh θ cosh θ r cosh θ r as reqired r d cosh θ cosh θ r sinh θ dθ dθ d dθ r r cosh θ sinh θ cosh θ sinh θ cosh θ sinh θ sinh θ cosech θ and so dθ sinh θ. We also have r coth θ cosech θ and cosh θ. d y r sinh θ dθ cosh θ r sinh θ cosh θ cosh θ r r STEP III Hyperbolics: Soltions 5

6 Differentiating again gives: d y d cosh θ r r d d cosh θ cosh θ r 4 r 4 sinh θ dθ cosh θ r r 4 r 4 sinh θ dθ cosh θ r cosech θ 4 sinh θ sinh θ r 4 sinh θ sinh θ + cosh θ cosh θ r 4 cosh θ r r cancelling r cosh θ cosh θ Looking at these reslts, a reasonable conjectre wold be dn y something cosh nθ n r n. To find a sitable expression for something, look back to see how these constants were formed previosly. It might be helpfl to look at what d4 y might be. If yo differentiated 4 again, the 4 wold be mltiplied by from both the power of r and the mltiple of θ. Hence we seem to have: n constant n constant n constant 4 n 4 constant So the something might be n!. Now we need to carry ot the proof by indction. Conjectre: dn y cosh nθ n!n n r n Base case: From the previos work, we can see that the conjectre is tre for n,,. STEP III Hyperbolics: Soltions 6

7 Indctive step: Assme the conjectre is tre when n k, so we have dk y cosh kθ k!k k r k. Differentiating with respect to x gives: d k+ y r k k sinh kθ dθ cosh kθ krk k!k k+ r k k r k! k k r k r sinh kθ dθ cosh kθ k k! k cosech θ sinh kθ sinh r k+ θ cosh kθ cosh θ k k! k sinh kθ sinh θ + cosh kθ cosh θ rk+ k coshk + θ k k! k coshk + θ k! r k+ r k+ Which is the same expression as the conjectre with n k +. Hence the conjectre is tre for n k then it is tre for n k +, and since it is tre for n it is tre for all integers n. STEP III Hyperbolics: Soltions 7

8 4 This is qite a long qestion! Sbstitting x a cosh T into the left hand side of the eqation gives: x a x 8a cosh T 6a cosh T a 4 cosh T cosh T a cosh T sing the first given reslt Hence x a cosh T is a soltion to the eqation. Comparing x bx c and x a x a cosh T is appears that we want to take b a which as b > b > is an ok thing to do. Frther we want c a cosh T i.e. cosh T c c. For this to be ok we need. We are a a told that c b, and as we are taking b a this means c a 6. As long as c and a have the same sign, this means that c a and c. a T We therefore know that one soltion is x a cosh Using the second reslt given at the start of the qestion we have: c T arcosh a c c ln a + a 6 c + c ln a 6 a. c + c ln b ln a ln a a Therefore the root becomes: T x a cosh a cosh ln a a e ln a + e ln a a a + a + a + b STEP III Hyperbolics: Soltions 8

9 We now have a root x + b, which means that x b is a factor of x bx c. There are varios ways to proceed, inclding long division or by sing: x bx c x b x + Ax + B Eqating coefficients for this last one gives s: B c + b A + b and so the other roots are the roots of the eqation: x + + b x + c + b. This is not in the reqired form yet, bt since x + b is a soltion to x bx c we have: c + b b + b + b + b b + b + b + b + b + b b b So now we know that the other roots are the roots of the eqation: x + + b x + + b b. Using the qaatic formla we have: + b ± + b + b ± + b + b + b ± + b + b + b ± ± i 4 + b b 4 4 b + 4b b b b STEP III Hyperbolics: Soltions 9

10 We now want this in terms of ω + i. We have: + b + i b + i + b i ω + b ω Noting that ω 4 i i. The other root is: + b i b i + b + i ω + b ω For the final part, we have x 6x 6 which means b and c. This gives: The soltions are therefore: a b c + c b b + b + + ω + b ω ω + ω ω + b ω ω + ω STEP III Hyperbolics: Soltions