Home » Blog » Prove some inequalities using the function ex-1-x

Prove some inequalities using the function ex-1-x

by RoRi
  1. Define a function:

        \[ f(x) = e^x - 1 - x \qquad \text{for all } x \in \mathbb{R}. \]

    Prove that f'(x) \geq 0 for all x \geq 0 and f'(x) \leq 0 for all x \leq 0. Prove the following inequalities are valid for all x > 0,

        \[ e^x > 1 + x, \qquad e^{-x} > 1 -x. \]

  2. Using integration and part (a) prove

        \[ e^x > 1 + x + \frac{x^2}{2!}, \qquad e^{-x} < 1 - x + \frac{x^2}{2!}. \]

  3. Using integration and part (a) prove

        \[ e^x > 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}, \qquad e^{-x} > 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!}. \]

  4. State and prove a generalization of the inequalities above.
  1. First, we take the derivative of f(x),

        \[ f'(x) = e^x - 1 \quad \implies \quad f'(0) = 0. \]

    Since e^x - 1 is strictly increasing everywhere (since f''(x) = e^x > 0 for all x) we have f'(x) \geq 0 for x \geq 0 and f'(x) \leq 0 for x \leq 0. Thus, f(x) has a minimum at x = 0 and f(0) = 0 so f(x) > 0 for all x \neq 0. Therefore, f(x) > 0 when x > 0; hence,

        \[ e^x -1 - x > 0 \quad \implies \quad e^x > 1 + x. \]

    Furthermore, when x < 0 we have

        \[ e^x - 1 - x > 0 \quad \implies \quad e^x > 1 + x. \]

    Therefore, if x > 0 then -x < 0 and so we have

        \[ e^{-x} > 1 - x. \]

  2. Using the inequalities in part (a) we integrate over the interval 0 to x,

        \begin{align*} e^x > 1 + x && \implies && \int_0^x e^t \, dt > \int_0^x (1+t) \, dt \\ && \implies && e^x - 1> \left(t + \frac{t^2}{2}\right)\Bigr \rvert_0^x + C \\ && \implies && e^x - 1> x + \frac{x^2}{2!} \\ && \implies && e^x > 1 + x + \frac{x^2}{2!} \end{align*}

    For the other inequality we proceed similarly,

        \begin{align*} e^{-x} > 1 - x && \implies && \int_0^x e^{-t} \, dt &> \int_0^x (1-t) \, dt \\ && \implies && -e^{-x} + 1 &> x - \frac{x^2}{2}\\ && \implies && e^{-x} &< 1 - x + \frac{x^2}{2!}. \qquad \blacksquare \end{align*}

  3. We use the same strategy as before, starting with the inequalities we established in part (b),

        \begin{align*} e^x > 1 + x + \frac{x^2}{2!} && \implies && \int_0^x e^t \, dt > \int_0^x \left (1 + t + \frac{t^2}{2!} \right) \, dt \\ && \implies && e^x - 1 > \left( t + \frac{t^2}{2!} + \frac{t^3}{3!}\right)\Bigr \rvert_0^x \\ && \implies && e^x > 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}. \end{align*}

    Similarly, for the other inequality,

        \begin{align*} e^{-x} < 1 - x + \frac{x^2}{2!} && \implies && \int_0^x e^{-t} \, dt &< \int_0^x \left( 1 - t + \frac{t^2}{2!} \right) \, dt \\ && \implies && -e^{-x} + 1& < x - \frac{x^2}{2!} + \frac{x^3}{3!} \\ && \implies && e^{-x} &> 1 - x +\frac{x^2}{2!} - \frac{x^3}{3!}. \qquad \blacksquare \end{align*}

  4. Claim: The following general inequalities hold for all x > 0:

        \begin{align*} e^x &> \sum_{k=0}^n \frac{x^k}{k!} & \text{for all } n \in \mathbb{Z}_{>0} \\ e^{-x} &> \sum_{k=0}^n (-1)^k \frac{x^k}{k!} & \text{for all odd } n \in \mathbb{Z}_{>0} \\ e^{-x} &< \sum_{k=0}^n (-1)^k \frac{x^k}{k!} & \text{for all even } n \in \mathbb{Z}_{>0}. \end{align*}

    Proof. The proof is by induction. We have already established the inequalities for the n = 1 case for all three inequalities. Now, to prove the first inequality assume

        \[ e^x > \sum_{k=0}^n \frac{x^k}{k!} \qquad \text{for some } n \in \mathbb{Z}_{>0}. \]

    Then we have

        \begin{align*} && \int_0^x e^t \, dt &> \int_0^x \left( \sum_{k=0}^n \frac{t^k}{k!} \right) \, dt \\[9pt] \implies && e^x - 1 &> \sum_{k=0}^n \int_0^x \frac{t^k}{k!} \, dt &(\text{linearity of integral}) \\[9pt] \implies && e^x &> 1 + \sum_{k=0}^n \left( \frac{t^{k+1}}{(k+1)!} \right)\Bigr \rvert_0^x \\[9pt] \implies && e^x &> 1 + \sum_{k=0}^n \frac{x^{k+1}}{(k+1)!} \\[9pt] \implies && e^x &> 1 + \sum_{k=1}^{n+1} \frac{x^k}{k!} &(\text{Reindexing})\\ \implies && e^x &> \sum_{k=0}^{n+1} \frac{x^k}{k!}. \end{align*}

    This establishes the first inequality. For the second inequality we have proved the case n = 1 already. Assume

        \[ e^{-x} > \sum_{k=0}^n (-1)^k \frac{x^k}{k!} \qquad \text{for some odd } n \in \mathbb{Z}_{>0}. \]

    Since n is odd we may write n = 2m+1 for some nonnegative integer m. We want to show that the inequality must hold for the next odd integer, n+2. We have,

        \begin{align*} &&\int_0^x e^{-t} \, dt&> \int_0^x \left( \sum_{k=0}^n (-1)^k \frac{t^k}{k!} \right) \, dt \\ \implies && -e^{-x} + 1 &> \sum_{k=0}^n (-1)^k \int_0^x \frac{t^k}{k!} \, dt \\ \implies && -e^{-x} &> -1 + \sum_{k=0}^n (-1)^k \frac{x^{k+1}}{(k+1)!} \\ \implies && e^{-x} &< 1 - \sum_{k=0}^n (-1)^k \frac{x^{k+1}}{(k+1)!} \\ \implies && e^{-x} &< 1 - \sum_{k=1}^{n+1} (-1)^{k-1} \frac{x^k}{k!} \\ \implies && e^{-x} &< 1 + \sum_{k=1}^{n+1} (-1)^k \frac{x^k}{k!} \\ \implies && e^{-x} &< \sum_{k=0}^{n+1} (-1)^k \frac{x^k}{k!} \\ \end{align*}

    We want to show that the inequality holds for n+2 so we integrate both sides again,

        \begin{align*} && e^{-x} &< \sum_{k=0}^{n+1} (-1)^k \frac{x^k}{k!} \\ \implies && \int_0^x e^{-t} \, dt &< \int_0^x \left( \sum_{k=0}^{n+1} (-1)^k \frac{x^k}{k!} \right) \, dx \\ \implies && -e^{-x} + 1 &< \sum_{k=0}^{n+1} (-1)^k \int_0^x \frac{t^k}{k!} \, dt \\ \implies && -e^{-x} &< -1 + \sum_{k=0}^{n+1} (-1)^k \frac{x^{k+1}}{(k+1)!} + C \\ \implies && -e^{-x} &< -1 + \sum_{k=1}^{n+2} (-1)^{k-1} \frac{x^k}{k!} \\ \implies && e^{-x} &> 1 - \sum_{k=1}^{n+2} (-1)^{k-1} \frac{x^k}{k!} \\ \implies && e^{-x} &> 1 + \sum_{k=1}^{n+2} (-1)^k \frac{x^k}{k!} \\ \implies && e^{-x} &> \sum_{k=0}^{n+2} (-1)^k \frac{x^k}{k!}. \end{align*}

    Hence, if the inequality holds for odd n, then it also holds for the next odd number, n+2. Hence, it holds for all positive, odd integers.
    The exact same induction argument works for all of the even integers (starting with the n = 2 case we proved in part (c)). \qquad \blacksquare

3 comments

  1. tom says:

    It’s not possible that e^x is strictly larger than 1 + x + x^2 because that implies e^0=1>1. I think Apostol made a mistake. Also, regarding the indefinite integral inequality, the RH integral which includes the constant C, if C is less then 1 how the inequality is still valid after adding more than C to the LH side? I was thinking of taking derivatives which would yield the original inequality, then since E(0) equals the polynomial @0 then things make sense but this seems like an inductive proof which seems out of context of the problem. Hopefully I’m not missing something here…

      • RoRi says:

        Hi. So, yes Apostol specified x > 0 so there is no error from him. I made a huge mistake in the approach to the problem. Taking indefinite integrals and looking at the value of the constant was just completely wrong. I’ve fixed it now… We take definite integrals from 0 to $x$ and evaluate everything and it works out. I’m sure I’ve introduced new errors in fixing the whole thing. (Basically all of parts (b) – (d) were wrong so I rewrote them completely.)

        (Also, I went ahead and incorporated your edits into your first comment and deleted the correction. Apparently, people cannot edit their own comments. I’ll try to work on the technology for that at some point, but definitely no promises on when that might get changed.)

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):