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Identify and explain why a statement is false

by RoRi

Given two statements:

  1. The integral

        \[ \int_{2 \pi}^{4 \pi} \frac{\sin t}{t} \, dt > 0 \]

    because

        \[ \int_{2 \pi}^{3 \pi} \frac{\sin t}{t} \, dt > \int_{3 \pi}^{4 \pi} \frac{|\sin t|}{t} \, dt. \]

  2. The integral

        \[ \int_{2 \pi}^{4 \pi} \frac{\sin t}{t} \, dt = 0 \]

    because, by the weighted mean value theorem (Theorem 3.16 in Apostol), there exists a c in [2 \pi, 4 \pi] such that

        \[ \int_{2 \pi}^{4 \pi} \frac{ \sin t}{t} \, dt = \frac{1}{c} \int_{2 \pi}^{4 \pi} \sin t \, dt = \frac{ \cos (2\pi) - \cos (4 \pi)}{c} = 0. \]

    3. Identify which of the two statements is false, and explain why.

Statement (b) is false since we may not apply the weighted mean value theorem in this case. The weighted mean value theorem requires the function g(t) to not change sign on the interval [2 \pi, 4 \pi]. Since g(t) = \sin t does change sign on the interval, we cannot apply the theorem.

Evaluate if some formulas for finite sums are true or false

by RoRi
    Decide whether the following are true or false:

  1. \displaystyle{\sum_{n=0}^{100} n^4 = \sum_{n=1}^{100} n^4}.
  2. \displaystyle{\sum_{j=0}^{100} 2 = 200}.
  3. \displaystyle{\sum_{k=0}^{200} (2+k) = 2 + \sum_{k=0}^{100} k}.
  4. \displaystyle{\sum_{i=1}^{100} (i+1)^2 = \sum_{i=0}^{99} i^2}.
  5. \displaystyle{\sum_{k=1}^{100} k^3 = \left(\sum_{k=1}^{100} k \right) \left( \sum_{k=1}^{100} k^2 \right)}.
  6. \displaystyle{\sum_{k=0}^{100} k^3} = \left( \sum_{k=0}^{100} \right)^3}.
  1. True. Since 0^4 = 0 the extra term on the left has no impact on the sum. In other words,

        \[ \sum_{n=0}^{100} n^4 = 0^4 + \sum_{n=1}^{100} n^4 = \sum_{n=1}^{100} n^4. \]

  2. False. We know from here that \sum_{j=1}^{100} 1 = 100; hence,

        \[ \sum_{j=0}^{100} 2 = 2 \left(1 + \sum_{j=1}^{100} 1 \right) = 202. \]

    (The “problem” here is the extra term in the sum. The j=0 term actually contributes to the sum since 2 evaluated at j=0 is still 2.)

  3. False. This is just abusing the additive property of finite sums. In reality,

        \[ \sum_{k=0}^{200} (2+k) = \left( \sum_{k=0}^{200} 2 \right) + \left( \sum_{k=0}^{200} k \right) = 202 + \sum_{k=0}^{100} k. \]

  4. False. This is reindexing the sum incorrectly. Seeing, (i+1)^2 the idea is to replace a sum over i, by a sum over i+1 to get a simpler expression. (A quick and dirty way to see that this has not been reindexed correctly is by looking at the first few terms. The sum on the left starts 2^2 + 3^2 + 4^2 + \cdots, while the sum on the right starts 0^2 + 1^2 + 2^2 + \cdots. To do this correctly, we replace i+1 by i in the interior, and then have to add one to the top and bottom indices so that we have i+1=2 to 101 and can replace the i+1 by i. This is the way I think about it; there are probably better ways.) Anyway, reindexing properly we would get,

        \[ \sum_{i=1}^{100} (i+1)^2 = \sum_{i=2}^{101} i^2. \]

    Or we could actually expand out the (i+1)^2 squared term and use linearity to separate the sums and evaluate all of them separately, but that would be slower.

  5. False. Sums and powers just don’t work this way. (This is really claiming (1^3 + 2^3 + \cdots + 100^3) = (1+2 + \cdots +100)(1^2 + 2^2 + \cdots + 100^2).) From work we did earlier (see: sums of integers, sums of squares, and sums of cubes) we have,

        \[ \sum_{k=1}^n k^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}, \]

    while

        \[ \left( \sum_{k=1}^n k \right) \left( \sum_{k=1}^n k^2 \right) = \left( \frac{n^2}{2} + \frac{n}{2} \right) \left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right). \]

    One can then plug in n=100 to see that we get 25502500 on the left, but 1708667500 on the right.

  6. False. Again, this just wrong. The claim is that (1^3 + 2^3 + \cdots + 100^3) = (1 + 2 + \cdots + 100)^3, which is of course false. More concretely, we know

        \[ \sum_{k=0}^{n} k^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}, \]

    while

        \[ \left( \sum_{k=0}^n k \right)^3 = \left( \frac{n^2}{2} + \frac{n}{2} \right)^3.\]

    So, evaluating at n=100, we have 25502500 on the right and 128787625000 on the right.

Find the error in an inductive “proof”

by RoRi

Consider the statement:

    \[ A(n): 1 + 2 + \cdots + n = \frac{1}{8} (2n+1)^2. \]

  1. Prove that if the statement is true for an integer k, then it is also true for k+1.
  2. What is wrong with the claim that this proves that A(n) is true for all n?
  3. Make a change to turn this into a true statement and prove it for all positive integers.
  1. Proof. Assume the statement is true for an integer k. Then, adding k+1 to both sides of the formula we assume is true for the case k,

        \begin{align*} 1+2+\cdots + k + (k+1) &= \frac{1}{8} (2k+1)^2 + (k+1) & (\text{Ind. Hyp.}) \\ &= \frac{k^2}{2} + \frac{k}{2} + \frak{1}{8} + k + 1 \\ &= \frac{4k^2 + 12k + 9}{8} \\ &= \frac{1}{8} (2(k+1)+1)^2 \end{align*}

    Thus, the statement is true for k+1 if it is true for k. \qquad \blacksquare

  2. Of course, the above is not a proof by induction, and does not show that the statement is true for all n since we have not shown that it is true for any base case. We have shown that if it is true for some n, then it is true for n+1. To show that it is true for all n, we must show that it is true for the case n=1 (or some other starting point). As one can calculate, it is not true for n=1, or any other n for that matter.
  3. In the case n=1 we have 1 on the left, and \frac{9}{8} on the right. So, we make the following:
    Claim:

        \[ A(n): 1 + 2 + \cdots + n < \frac{1}{8} (2n+1)^2. \]

    Proof. We have already shown it is true for n=1 since 1 < \frac{9}{8}. Now assume it is true for some integer k, then using the inductive hypothesis and adding k+1 to both sides we have

        \begin{align*} 1 +2 + \cdots + k + (k+1) &< \frac{1}{8} (2k+1)^2 + (k+1) \\ &< \frac{1}{8} (4k^2 + 12k + 9) \\ &< \frac{1}{8} (2(k+1)+1)^2. \end{align*}

    Thus, the inequality is true for all n. \qquad \blacksquare