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Y=(e^ax)(sinbx) - Show that d^ny/dx^n=

  1. Jul 14, 2009 #1
    1. The problem statement, all variables and given/known data
    T=θ (it won't let me display theta inside the "tex" boxes.)

    Show that: [tex]\frac{d^n}{dx^n}\right) e^a^x(sinbx) = r^n*e^a^x*sin(bx+nT)[/tex]
    where a and b are positive numbers, [tex]r^2=a^2+b^2[/tex],

    and [tex]T=tan^-^1(\frac{b}{a}\right))[/tex].


    2. Relevant equations
    Trigonometric functions.
    [tex]sinT=(\frac{b}{r}\right))[/tex]
    etc...


    3. The attempt at a solution
    Since [tex]r^2=a^2+b^2[/tex]; I've constructed a circle with radius r with a right triangle constructed inside the circle whose opposite = b, adjacent = a, and hypotenuse = r.

    [tex]y=e^a^x*sinbx[/tex]

    From differentiating with respect to x, I get:
    [tex]y'=e^a^x(a*sin(bx)+b*cos(bx))[/tex]

    From the trigonometric functions, I've explicitly defined a and b:
    [tex]b=r*sinT[/tex], [tex]a=r*cosT[/tex]

    By directly substituting the values of a and b into y', I get:
    [tex]y'=e^a^x(r*sin(bx)cos(T)+r*cos(bx)sin(T))[/tex]

    I've factored r from and used the trigonometric addition and subtraction formula to simplify the term inside the parentheses. Now I have:
    [tex]y'=r*e^a^x(sin(bx+T))[/tex]

    Here is where I'm a bit confused as how the show [tex]y'=r^n*e^a^x(sin(bx+nT))[/tex]. Specifically the terms "[tex]r^n[/tex]" and "[tex]nT[/tex]".

    I've tried differentiating [tex]\frac{d^ny}{dx^n}\right)[/tex], but that leads to an [tex]nx^n^-^1[/tex] term and doesn't address how I could get [tex]nT[/tex] in my final expression.
     
  2. jcsd
  3. Jul 14, 2009 #2
    One way to do this problem is to use the method of mathematical induction: assume that the proposition holds for some n = k and prove that it follows from this for n = k + 1. This method is quite simple...try it.

    The other way to do it is using Euler's Identity:

    [tex]e^{i\alpha} = \cos\alpha + i\sin\alpha[/itex]

    where [itex]i = \sqrt{-1}[/itex].

    Using this, one can express [itex]\sin bx[/itex] as [itex]\frac{e^{ibx}-e^{-ibx}}{2i}[/itex]. So

    [tex]e^{ax}\sin bx = \frac{1}{2i}\left[e^{(a+ib)x}-e^{(a-ib)x}\right][/tex]

    Now, one can write

    [tex]a + ib = r e^{i\theta}[/tex]

    where [itex]r = \sqrt{a^2+b^2}[/itex] and [itex]\theta = \tan^{-1}\frac{b}{a}[/itex].

    and use [itex]\frac{d^{n}}{dx^n}e^{kx} = k^{n}x[/itex] for real and complex [itex]k[/itex].

    I have skipped a lot of steps here, assuming you can fill in the details. I have also assumed that you know something about complex numbers.
     
  4. Jul 15, 2009 #3
    Well I just finished differential calculus so I have no idea how to do anything with complex numbers but i've tried proving the statement with mathematical induction but am a bit confused how to justify last statement.

    Here's what I got:

    T=θ

    General term: [tex]a_n[/tex] = [tex]\frac{d^ny}{dx^n}\right)[/tex]

    [tex]a_1 = S_1[/tex]

    [tex]\frac{d^1y}{dx^1}\right)=r^1*e^a^xsin(bx+(1)T)[/tex]

    Which simplifies to:
    [tex]\frac{d^1y}{dx^1}\right)=r^*e^a^xsin(bx+T)[/tex]

    Thus:
    [tex]\frac{d^1y}{dx^1}\right)=\frac{dy}{dx}\right)[/tex]

    Thus, the statement [tex]S_1[/tex] is true, therefore, i made the assumption that the statement [tex]S_k[/tex] is true aswell.

    Therefore (to my understanding), I need to prove:

    [tex]S_(_k_+_1_)=S_k[/tex]

    Thus:

    [tex]S_(_k_+_1_) = \frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^xsin(bx+kT+T)[/tex]

    I've grouped the terms bx+kT and expanded the term sin([bx+kT]+T) using the trignonometric addition and subtraction formula.

    [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^x[sin(bx+kT)cos(T)+cos(bx+kT)sin(T)][/tex]

    From the trig +/- formula, again...:

    [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^x[sin(bx)cos(kT)cos(T)+cos(bx)sin(kT)cos(T)+cos(bx)cos(kT)sin(T)-sin(bx)sin(kT)sin(T)][/tex]

    Here's where I can't prove the induction, from grouping terms with like-terms, it simplifies back to [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = ...[/tex] with the terms inside the sine re-arranged which doesn't even matter since you can do that anyways since bx, kT and T are all positive.

    This may seem silly that I don't know how to use mathematical induction but a lot of the earlier theorems, axioms, etc... before pre-calc2: trig. i'm pretty unclear on. ps disclaimer: i based my method of an instructional video from youtube, wikipedia, and some other sites demonstrating induction.

    EDIT:

    alright so i've managed to crank out a solution that seems pretty conclusive, however, i'm not sure if the steps taken to reach the answer are allowed. here's what i've done...

    starting from:

    [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^x[sin(bx)cos(kT)cos(T)+cos(bx)sin(kT)cos(T)+cos(bx)cos(kT)sin(T)-sin(bx)sin(kT)sin(T)][/tex]

    From the trigonometric functions, i've substituted the value of θ into the expression [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right)[/tex]:

    [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^x[sin(bx)cos(kcos^-^1(\frac{a}{r}\right)))cos(cos^-^1(\frac{a}{r}\right)))+cos(bx)cos(kcos^-^1(\frac{a}{r}\right)))sin(sin^-^1(\frac{b}{r}\right)))+cos(bx)sin(ksin^-^1(\frac{b}{r}\right)))cos(cos^-^1(\frac{a}{r}\right)))-sin(bx)[/tex]
    [tex]sin(ksin^-^1(\frac{b}{r}\right)))sin(sin^-^1(\frac{b}{r}\right)))][/tex]

    Here's where i'm unsure if this operation is allowed:

    [tex]sin(ksin^-^1(\frac{b}{r}\right))) = sink(sin^-^1(\frac{b}{r}\right)))[/tex] ?

    Assuming it is, I used the rule [tex]sin(sin^-^1(x))=x[/tex] to re-express the expression:

    [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^x[sin(bx)(\frac{a^2k}{r^2}\right)-\frac{b^2k}{r^2}\right))+cos(bx)(2*\frac{abk}{r^2}\right))][/tex]

    From the trigonometric functions, I substituted the values of the terms [tex]\frac{a^2}{r^2}\right)[/tex], [tex]\frac{b^2}{r^2}\right)[/tex], [tex]\frac{2ab}{r^2}\right)[/tex] into the expression and factored k. Thus i have:

    [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^x[sin(bx)*k*(cos^2(T)-sin^2(T))+cosbx(2kcos(T)sin(T))][/tex]

    Using the double angle formula, i get:

    [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^x[sin(bx)cos(2kT)+cos(bx)sin(2kT)][/tex]

    ...and the double-angle formula gives:

    [tex]\frac{d^(^k^+^1^)y}{dx^(^k^+^1^)}\right) = r^(^k^+^1^)*e^a^xsin(bx+2kT)[/tex]

    which I concluded for any higher-order derivative k, the next derivative (k+1) would look like the above expression since sin2T=2sinTcosT and sin4T=4sinTcosT, sin2kT=2ksinTcosT (where k >0), etc... which is conclusive reasoning to prove the statement.

    [tex]\frac{d^n}{dx^n}\right)(e^axsin(bx)) = r^ne^axsin(bx+nT)[/tex]
     
    Last edited: Jul 15, 2009
  5. Jul 15, 2009 #4
    Greetings, and I must first commend you on working out the details carefully and typing them out here :-)

    No problem, I'll give you a quick primer in this post before looking at the second part of your solution in my second reply (in a short while).

    The idea in mathematical induction is to prove a proposition P(n + 1) (n = a natural number) based on the assumption that P(n) is true.

    The process is three fold:

    1. Basis Step: Prove P(n) for some small value(s) of n, say n = 1.
    2. Inductive Hypothesis: Assume the validity of P(n), i.e. assume P(n) is true.
    3. Prove that the truth of P(n+1) must follow from 1 and 2.

    There will be some minor variations of this method, depending on the nature of the proposition P(n). In your problem,

    [tex]P(n): \frac{d^n}{dx^n}\right) e^a^x(sinbx) = r^{n}e^{ax}sin(bx+nT)[/tex]

    For n = 1, evaluate the LHS and RHS separately, show that they are equal. Then assume that P(n) is true.

    Now, can you think of a way to get the LHS of P(n+1) by operating on (read: doing something to) the LHS of P(n)? Work it out..
     
    Last edited: Jul 15, 2009
  6. Jul 15, 2009 #5
    In your first method, you have used induction separately on the LHS and RHS. Remember: induction is on a proposition, not on an expression. So you have to consider the entire proposition

    In your notation, let [itex]a_{n}[/itex] and [itex]S_{n}[/itex] denote the LHS and RHS separately. You have to prove that

    [tex]a_{n} = S_{n}[/tex]

    for all n.

    As a general rule, if the RHS you are aiming to get is compact, you shouldn't expand it in the middle of an inductive step, as this will only make the expression more intractable.

    What you are claiming here is

    [tex]\sin(k\sin^{-1}x) = \sin^{-1}(k\sin x)[/tex]

    which is clearly false. Note that this is wrong even for k = 1 (can you tell me why?).

    (Hint: consider x in the interval [itex](\pi/2, \pi)[/itex] and look at the LHS and RHS for k = 1.)
     
  7. Jul 16, 2009 #6
    Alright, here's my induction:

    Statement: [tex]P_n = S_n[/tex] for all natural numbers n. (n=k)

    [tex]P_n: \frac{d^n}{dx^n} (e^a^xsinbx)[/tex]

    [tex]S_n: r^ne^a^xsin(bx+nT)[/tex]

    (T=θ)

    1) [tex]P_1[/tex] holds true. (n=1)

    [tex]P_1: \frac{d^n}{dx^n} (e^a^xsinbx) = re^a^xsin(bx+T)[/tex]

    [tex]S_1: r^ne^a^xsin(bx+nT) = r^(^1^)e^a^xsin(bx+(1)T) = re^a^xsin(bx+T)[/tex]

    Thus, for n=1, [tex] P_1 = S_1[/tex]

    2) [tex]P_k[/tex] is assumed to be true for all natural numbers k. (n=k)

    3) Thus, if [tex]P_k[/tex] is true, then [tex]P_(_k_+_1_)[/tex] is also true.
    Therefore: k < k+1
    And: m = k+1
    Thus: k < m

    For the the statement [tex]P_k[/tex] is true for k ≥ 1,
    And if k < m,
    Then m > 1.

    If [tex]P_k[/tex] is true for k ≥ 1, then it is also true for k > 1 therefore, [tex]P_m[/tex] is true aswell.
    And if m = k + 1, then [tex]P_(_k_+_1_)[/tex] is true.

    If [tex]P_n[/tex] is true for all natural numbers n (n=1, k, k+1) and all three cases hold true, therefore the statement [tex]P_n = S_n[/tex] is true. Thus proving the statement:

    [tex]\frac{d^n}{dx^n} (e^a^xsinbx) = r^ne^a^xsin(bx+nT)[/tex]

    QED? :confused:
     
  8. Jul 16, 2009 #7

    rock.freak667

    User Avatar
    Homework Helper

    Prove:
    [tex]\frac{d^n}{dx^n}\right) e^a^x(sinbx) = r^n*e^a^x*sin(bx+nT) [/tex]

    Assume true for n=k


    [tex]\frac{d^k}{dx^k}\right) e^a^x(sinbx) = r^k*e^a^x*sin(bx+kT) [/tex]

    You proved true for k=1 already.

    To prove true for n=k+1, differentiate both sides w.r.t. x

    so that


    [tex]\frac{d}{dx}(\frac{d^k}{dx^k}\right) e^a^x(sinbx)) = \frac{d}{dx}(r^k*e^a^x*sin(bx+kT)) [/tex]

    Now when you differentiate the right side side use this fact,

    AsinX+BcosX =R sin(X+C)
    = R sinXcosC + RsinCcosX

    such that

    RcosC=A
    RsinC=B

    so that R2=A2+B2

    and that [itex]tan^{-1}(\frac{B}{A})=C[/itex]

    Just use that and you will see that it works out very easily.
     
  9. Jul 18, 2009 #8
    Samuel, hopefully you got it...rock.freak67 has given you ALL the steps. You just need to work it out. As for your post, I would like to point out that you want to prove

    P(n) = S(n) (in your notation)

    for all n.

    So, you first prove P(1) = S(1).

    Then you assume P(k) = S(k) and differentiate both sides and simplify to get P(k+1) = S(k+1).

    Then you assert that by the principle of mathematical induction, since the truth of P(k+1) = S(k+1) follows from the truth of P(k) = S(k), therefore P(n) = S(n) holds for all n.

    Note: Don't say P(1) is true. That means nothing: P(1) is an expression. As I stated in my post, induction works on a proposition, i.e. a mathematical statement which is expressed in terms of an equality of two quantities, the LHS and RHS. Induction helps you prove that LHS is always equal to the RHS for all natural numbers n.
     
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