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Questions To Be Studied in Differential Equations

  1. (12/22/2011) We can prove |sin z| £ sinh |z| using Taylor series. In my opinion, one cannot prove this inequality using other methods. In the inequality the arithmetic mean ³ the harmonic mean, one can insert the geometric mean in between. In contrast, the inequality |sin z| £ sinh |z| is tight, one cannot insert anything in between.
    Example. Suppose the power series S an t n is absolutely convergent in the neighborhood of t = 0. If |S an t n| £ S bn |t| n, where 0£ bn£ |an| and t is in a neighborhood of 0, then bn = |an|.
    Partial solution. For the Taylor series of sin t, let t = ix.


  2. (6/29/2012) Let f be an entire function. Suppose f has an essential singularity at z = ¥ [Guo, p.348, l.9]. If f (an) = g (an), where an®¥. Find all the methods of proving f (bn) = g (bn), where bn®¥.
    Example 1. [Guo, p.351, l.3-l.8]
    Example 2. [Guo, p.362, (15); [1]]


  3. (9/19/2012) Designing devices for obtaining direct formulas
    Let cos 2m z = Sn=0m amn cos 2nz. We may derive the recursion formula for amn by using the following two formulas:
    cos 2m z = cos 2 z cos 2m-2 z; cos 2 z = [cos 2z +1]/2. We may also obtain a direct formula for amn by using
    cosn x = 2-n (eix + e-ix) n. Recursion formulas have shortcomings. For example, in [Inc1, p.175, l.-7] we have recursion formula for cr, but we cannot use the recursion formula to prove the formula given in [Inc1, p.176, l.-17]. This is because the recursion formula can lower the precision, but cannot raise precision. Thus, Mathieu designed an integral equation given in [Wat1, p.407, l.-6]  as a device to directly obtain the qr+2k-term in the expansion of A2r(q), where A2r(q) satisfies
    ce0(z, q) = Sr=0¥ A2r(q) cos 2rz. [1] [Guo, p.630, l.2-l.14; p.630, l.15-p.632, l.10] give two more devices.
    Are there any other useful devices to transform a recursion formula to a direct formula?.


  4. (11/24/2012) Reduce the total length of the argument of reduction to absurdity given in [Perr, p.18, l.-13-p.19, l.6].


  5. (12/1/2012) (Minimum proof paths)
    Example 1. If A1, A2, A3 are equivalent, it suffices to show A1ÞA2ÞA3ÞA1, which is a minimum proof path. By simple logic, it is unnecessary to prove A1ÞA3.
    Example 1'. In Example 1, we use the following method of deduction: if (AÞB and BÞC), then AÞC. Suppose in another world people use a different method of deduction: If (A®B and A®C), then B«C. Now suppose A1, A2, A3 are equivalent again. Find a minimum proof path.
    Example 2. If (Ai and Aj)ÞAi+j (mod n), where (i, j Î{1,2,…,n} and i¹j), then no proofs can be omitted. For the case n=3, see [Perr, p.22, Satz 1].
    Formulate a statement for symmetric cases in graph theory that can cover the general situation and then prove it.


  6. (1/8/2013) (The linear independence for integral solutions)
        We want to prove that the two integrals ò[0, 1] e-xt f(t) dt and ò[1, +¥) e-xt f(t) dt are linearly independent [Inc1, p.189, l.14].
    Proof. Assume they are linearly dependent: ò[0, 1] e-xt f(t) dt = A ò[1, +¥) e-xt f(t) dt.
    Let F(x, t) be e-xt f(t) on tÎ[0, 1) and -A e-xt f(t) on tÎ[1, +¥).
    Then we have ò[0, +¥) F(x, t) dt = 0. By repeated differentiations with respect to x, we have
    ò[0, +¥) tn F(x, t) dt = 0. Then for every Borel set E, we  have òE F(x, t) dt = 0.
    Consequently, F(x, t) = 0 a.e. for tÎ[0, +¥), a contradiction.
    Thus, we have handled the case given in [Guo, p.80, (16)]. Similarly, we can deal with the cases given in [Guo, p.80, (17) & (18)]. Thus, we can prove the two integrals along the two contours given in the figure in [Wat1, p.307] are linearly independent: replace [0, 1) and [1, +¥) with the two contours, replace the Laplace transform with the Euler transform and note that the reciprocal of a continuous function is continuous. How do we handle the above cases for ODE's of nth order? Consider the disjoint union of the n contours of integration. How do we treat the general case?


  7. (9/30/2014) The equality given in [Wat, p.36, l.-3] provides a resource to study the conditions under which a term-by-term integration preserves a series' uniform convergence.


  8. (10/19/2014) We may generalize Euler’s solutions to the two differential equations given in [Wat, p.62, l.-8-l.-2] as follows:
    Let y=x1/2u and z=2na1/2x1/(2n). Then the following two differential equations are equivalent:
    x3/2(d2y)/(d2x)+ax(n-2)/(2n)y=0;
    z2(d2u)/(d2z)+z(du/dz)+(z2-n2)u=0.
    Given an arbitrary differential equation of the second order. Can we use similar fixed transformations to obtain an equivalent differential equation of the form xa(d2y)/(d2x)+ay=0, where a is a rational number?


  9. (2/16/2015) I suspect that the construction given in [Fin, §97] and that given in [Fin, §217] are dual. In other words, if we make small changes in rules, they may still produce the same figure. The evidences of duality are given as follows: Let Figure 1 be the figure given in [Fin, p.74] and Figure 2 be the first figure given in [Fin, p.178]. Then
    F, F' in Figure 1 are fixed points; x-axis and y-axis in Figure 2 are fixed lines.
    P in Figure 1 is a moving point connected to fixed points by a line segment whose total length is PF+PF'= 2a. AB in Figure 2 is a moving line connected to fixed lines; AB is divided by a point P such that PB=a and PA=b.


  10. (4/11/2015) Given a system of equations as in [Bel63, p.216, (1), (2) & (3)]. Let r be the rank of 3´3 coefficient matrix and r' be the 3´4 argumented matrix. The classification of conicoids is given by [Fin, p.283, Table]. [Bel63, p.220, l.21-l.22] says that (r=2 and r'=3) Û(the conicoid is a paraboloid)]. [Bel63, §158] says that (r=1 and r'=2) Û(the conicoid is a parabolic cylinder)]. [Bel63, §159] says that (r=r'=1) Û(the conicoid is a pair of parallel planes)]. Thus, the classification by ranks of submatrices of the matrix given in [Fin, p.266, (6)] is finer than the classification given in  [Fin, p.283, Table]. Can we use the ranks of submatrices of a (n+2)´(n+2) symmetric matrix to classify the general surfaces of degree n?


  11. (1/4/2017) Why does a series representation have an advantage over an integral representation in calculation? [Leb, §9.5]