2.5: Exercises
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Exercise 2.5.1
Show that if a function is differentiable, then it is also continuous.
Exercise 2.5.2
Prove that the derivative of ln(x) is 1/x.
- Answer
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If y=ln(x), it follows from the definition of the logarithm that exp(y)=x. Taking d/dx on both sides, and using the product rule, gives dydxexp(y)=1⇒dydx=1exp(y)=1x.
Exercise 2.5.3
Using the definition of non-natural powers, prove that ddx[xy]=yxy−1,forx∈R+,y∉N.
Exercise 2.5.4
Consider f(x)=tanh(αx).
- Sketch f(x) versus x, for two cases: (i) α=1 and (ii) α≫1.
- Sketch the derivative function f′(x) for the two cases, based on your sketches in part (A) (i.e., without evaluating the derivative directly).
- Evaluate the derivative function, and verify that the result matches your sketches in part (B).
Exercise 2.5.5
Prove geometrically that the derivatives of the sine and cosine functions are: ddxsin(x)=cos(x),ddxcos(x)=−sin(x). Hence, derive their Taylor series, Eqs. (2.2.5) and (2.2.6).
Exercise 2.5.6
For each of the following functions, derive the Taylor series around x=0:
- f(x)=ln[αcos(x)], to the first 3 non-vanishing terms.
- f(x)=cos[πexp(x)], to the first 4 non-vanishing terms.
- f(x)=1√1±x, to the first 4 non-vanishing terms. Keep track of the signs (i.e., ± versus ∓).
Exercise 2.5.7
For each of the following functions, sketch the graph and state the domains over which the function is differentiable:
- f(x)=|sin(x)|
- f(x)=[tan(x)]2
- f(x)=11−x2
Exercise 2.5.8
Let →v(x) be a vectorial function which takes an input x (a number), and gives an output value →v that is a 2-component vector. The derivative of this vectorial function is defined in terms of the derivatives of each vector component: →v(x)=[v1(x)v2(x)]⇒d→vdx=[dv1/dxdv2/dx]. Now suppose →v(x) obeys the vectorial differential equation d→vdx=A→v, where A=[A11A12A21A22] is a matrix that has two distinct real eigenvectors with real eigenvalues.
- How many independent numbers do we need to specify for the general solution?
- Let →u be one of the eigenvectors of A, with eigenvalue λ: A→u=λ→u. Show that →v(x)=→ueλx is a specific solution to the vectorial differential equation. Hence, find the general solution.
- Answer
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For an ordinary differential equation for a scalar (one-component) function of order n, the general solution must contain n independent variables. In this case, →v is a two-component function, so it requires 2n indpendent variables. The differential equation d→vdx=A→v has order n=1, so a total of 2 independent variables is required for the general solution.
Let u be an eigenvector of A with eigenvalue λ, and suppose that →v(x)=→ueλx (note that →u itself does not depend on x). Then d→vdx=→uddx(eλx)=λ→ueλx=(A→u)eλx=A(→ueλx)=A→v(x). Hence, →v(x) satisfies the desired differential equation.
Let →u1 and →u2 be the eigenvectors of A, with eigenvalues λ1 and λ2. The general solutions will be →v(x)=c1→u1eλ1x+c2→u2eλ2x, where c1 and c2 are independent variables.