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- (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*a*_{n }t^{n}is absolutely convergent in the neighborhood of t = 0. If |S*a*_{n }t^{n}| £ S*b*_{n}|t|^{n}, where 0£*b*_{n}£ |*a*_{n}| and_{ }t is in a neighborhood of 0, then*b*_{n}= |*a*_{n}|.

Partial solution. For the Taylor series of sin t, let t = ix. - (6/29/2012) Let f be an entire function. Suppose f has an essential singularity at z =
¥ [Guo, p.348, l.9]. If f (
*a*_{n}) = g (*a*_{n}), where*a*_{n}®¥. Find all the methods of proving f (b_{n}) = g (b_{n}), where b_{n}®¥.

Example 1. [Guo, p.351, l.3-l.8]

Example 2. [Guo, p.362, (15); [1]] - (9/19/2012) Designing devices for obtaining
*direct*formulas

Let cos^{2m}z = S_{n=0}^{m}a_{mn}cos 2nz. We may derive the*recursion*formula for a_{mn}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 a_{mn}by using

cos^{n }x = 2^{-n }(e^{ix }+ e^{-ix})^{ n}. Recursion formulas have shortcomings. For example, in [Inc1, p.175, l.-7] we have recursion formula for c_{r}, 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 q^{r+2k}-term in the expansion of A_{2r}(q), where A_{2r}(q) satisfies

ce_{0}(z, q) = S_{r=0}^{¥ }A_{2r}(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?. - (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].
- (12/1/2012) (Minimum proof paths)

Example 1. If A_{1}, A_{2}, A_{3}are equivalent, it suffices to show A_{1}ÞA_{2}ÞA_{3}ÞA_{1}, which is a minimum proof path. By simple logic, it is unnecessary to prove A_{1}ÞA_{3}.

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 A_{1}, A_{2}, A_{3}are equivalent again. Find a minimum proof path.

Example 2. If (A_{i}and A_{j})ÞA_{i+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. - (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, +¥)}t^{n}F(x, t) dt = 0. Then for every Borel set E, we have ò_{E}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? - (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.
- (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=x^{1/2}u and z=2n*a*^{1/2}x^{1/(2n)}. Then the following two differential equations are equivalent:

x^{3/2}(d^{2}y)/(d^{2}x)+*a*x^{(n-2)/(2n)}y=0;

z^{2}(d^{2}u)/(d^{2}z)+z(du/dz)+(z^{2}-n^{2})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 x^{a}(d^{2}y)/(d^{2}x)+*a*y=0, where a is a rational number? - (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'= 2*a*. 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. - (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?
- (1/4/2017) Why does a series representation have an advantage over an integral representation in calculation? [Leb, §9.5]
- (12/20/2018) Try to make the proof of [Car, p.37, Theorem 2] more straightforward.