A Note on Double Limit Problems

Excerpts

• A Note on Double Limit Problems

  Another way of expressing this fact is to say that the operations of summation from $0$ to $\infty$, and of integration from $0$ to $x$, are *commutative* when applied to the function $(-1)^{n}t^{n}$, *i.e.* that it does not matter in what order they are performed on the function.

• A Note on Double Limit Problems

  We can always, by the exercise of a little ingenuity, find $z$ so that $LL'z$ and $L'Lz$ shall differ from one another.

• A Note on Double Limit Problems

  Of course, in an exact science like pure mathematics, we cannot be satisfied with an answer of this kind; and in the higher branches of mathematics the detailed investigation of these questions is an absolute necessity.

• A Note on Double Limit Problems

  The operations of proceeding to the limit zero with each of two variables $x$ and $y$ may or may not be commutative when applied to a function $f(x, y)$.

• A Note on Double Limit Problems

  It is natural to suppose so: but that is all that we have a right to say at present.

• A Note on Double Limit Problems

  The preceding examples suggest that there are three possibilities with respect to the commutation of two given operations, viz.: (1) the operations may *always* be commutative; (2) they may *never* be commutative, *except in very special circumstances*; (3) they may be commutative *in most of the ordinary cases which occur practically*.

• A Note on Double Limit Problems

  In practice, a result obtained by assuming that two limit-operations are commutative is *probably* true: it at any rate affords a valuable *suggestion* as to the answer to the problem under consideration.

• A Note on Double Limit Problems

  Detailed investigations of a large number of important double limit problems will be found in Bromwich’s *Infinite Series*.

Equations

• A Note on Double Limit Problems

  -1 < x \leq 1

  The range of x for which the logarithmic series holds, as proved in section 213.

• A Note on Double Limit Problems

  \log(1 + x) = x - \tfrac{1}{2}x^{2} + \tfrac{1}{3}x^{3} - \dots

  The logarithm of 1 + x equals an infinite alternating series in x, valid for -1 < x ≤ 1.

• A Note on Double Limit Problems

  1/(1 + t) = 1 - t + t^{2} - \dots

  The function 1/(1 + t) is expanded as an infinite series in powers of t.

• A Note on Double Limit Problems

  \int_{0}^{x} \frac{dt}{1 + t} = \int_{0}^{x} dt - \int_{0}^{x} t\, dt + \int_{0}^{x} t^{2}\, dt - \dots

  Integrating the series term by term between 0 and x gives the integral of 1/(1 + t) as the sum of the integrals of its terms.

• This equation is in THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS \\ OF A REAL VARIABLE (THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS \\ OF A REAL VARIABLE)

• A Note on Double Limit Problems

  D_{x} \left(1 + x + \frac{x^{2}}{2!} + \dots\right) = D_{x}1 + D_{x}x + D_{x} \frac{x^{2}}{2!} + \dots

  The derivative of the exponential series equals the series of the derivatives of its terms, so the operations of differentiation and summation commute here.

• A Note on Double Limit Problems

  \lim_{x\to\xi} \left(1 + x + \frac{x^{2}}{2!} + \dots\right) = 1 + \xi + \frac{\xi^{2}}{2!} + \dots = \lim_{x\to\xi} 1 + \lim_{x\to\xi} x + \lim_{x\to\xi} \frac{x^{2}}{2!} + \dots

  The limit of the exponential series as x tends to ξ equals the sum of the termwise limits, so exp x is continuous at x = ξ.

• A Note on Double Limit Problems

  \lim_{x\to 0} \{\lim_{y\to 0} (x + y)\} = \lim_{x\to 0} x = 0

  Taking the limit in y first and then in x gives 0 for the function x + y.

• A Note on Double Limit Problems

  \lim_{y\to 0} \{\lim_{x\to 0} (x + y)\} = \lim_{y\to 0} y = 0

  Taking the limits in the opposite order also gives 0 for x + y.

• A Note on Double Limit Problems

  \lim_{x\to 0} \left(\lim_{y\to 0} \frac{x - y}{x + y}\right) &= \lim_{x\to 0} \frac{x}{x} &&= \lim_{x\to 0} 1 = 1

  Taking the limit in y first and then in x gives 1 for (x - y)/(x + y).

• A Note on Double Limit Problems

  \lim_{y\to 0} \left(\lim_{x\to 0} \frac{x - y}{x + y}\right) &= \lim_{y\to 0}\frac{-y}{y} &&= \lim_{y\to 0} (-1) = -1

  Taking the limit in x first and then in y gives -1 for (x - y)/(x + y), so the two iterated limits differ.

• A Note on Double Limit Problems

  \lim_{x\to 1} \left\{\sum_{1}^{\infty} \frac{(-1)^{n}}{n}x^{n}\right\} &= \lim_{x\to 1}\log(1 + x) &&= \log 2

  Summing the series first and then letting x tend to 1 from below gives log 2.

• A Note on Double Limit Problems

  \sum_{1}^{\infty} \left\{\lim_{x\to 1} \frac{(-1)^{n}}{n}x^{n}\right\} &= \quad \sum_{1}^{\infty} \frac{(-1)^{n}}{n} &&= \log 2

  Taking the limit of each term first and then summing also gives log 2 in this case, so the operations commute here.

• A Note on Double Limit Problems

  \lim_{x\to 1} \left\{\sum_{1}^{\infty} (x^{n} - x^{n+1})\right\} &= \lim_{x\to 1} \{(1 - x) + (x - x^{2}) + \dots\}

  Summing the telescoping series first and then taking the limit as x tends to 1 is set up as an iterated limit.

• A Note on Double Limit Problems

  \lim_{x\to 1} \{(1 - x) + (x - x^{2}) + \dots\} = \lim_{x\to 1} 1 = 1

  The telescoping series sums to 1 for x < 1, so its limit as x tends to 1 is 1.

• A Note on Double Limit Problems

  \sum_{1}^{\infty} \left\{\lim_{x\to 1} (x^{n} - x^{n+1})\right\} &= \sum_{1}^{\infty} (1 - 1) = 0 + 0 + 0 + \dots = 0

  Taking the limit of each term first and then summing gives 0, which differs from the limit of the sum (1), so these operations do not commute.

Problems

No exercises in this chapter.