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My professor told me to use Taylor Theorem with remainder form to establish this in equality but I don't get how to use this theorem. Need some help Thank you

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    Is my edit to your title correct? – 2017-02-25
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    @mrnovice yeah, thanks...did u get my question? Any hint for solving this? – 2017-02-25
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    Yes, just provide me with some context first, do you understand what a Taylor's series is, and how to derive it? – 2017-02-25
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    Also if this equality is true, it can only hold if it's $x-x^{2} \leq log(1+x) \leq x - \frac{x^{3}}{2(1+x)}$ – 2017-02-25
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    @mrnovice yeah, I'm completely known to taylor series derivation but don't know how to apply remainder form to establish this inequality. Choosing f(x) = log(1+x) – 2017-02-25
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    @Rahul What do you think about my solution? – 2017-02-25
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    @Rahul: You *removed* the restriction $0 \le x \le 1$ from the question title. But actually the inequality is wrong for some $x > 1$ (as pointed out below) and also wrong for $x < 0$. – 2017-02-27
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    @MartinR The $0 \le x \le 1$ constraint was edited into the (title of the) question by someone *other* than the OP (per the edit history). Quoting from an earlier comment of mine: *note that I am not saying that the condition doesn't makes sense, but only that it was never mentioned by the OP to begin with*. – 2017-02-27
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    @dxiv: Yes I know and agree. I just wondered that OP himself removed the condition instead of confirming it and clarifying the question. – 2017-02-27
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    @MichaelRozenberg: Proof for the upper bound: $\displaystyle x-\log(1+x)=\int_0^t \frac{t}{1+t}dt>\frac1{1+x}\int_0^x tdt=\frac{x^2}{2(1+x)}$. $\forall x\ge0$. Moreover, $\displaystyle\frac{x^2}{2(1+x)}>\frac{x^3}{2(1+x)}$, $\forall x\in(0,1)$. – 2017-03-04

2 Answers 2

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$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$.

Thus, $\ln(1+x)\geq x-\frac{x^2}{2}$ and

$$\ln(1+x)\leq x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}-\frac{x^8}{8}+\frac{x^9}{9}\leq x-\frac{x^3}{2(1+x)},$$ where the last inequality it's $$(1260-840x-210x^2+126x^3-84x^4+60x^5-45x^6+37x^7-280x^8)x^2\geq0,$$ which is obvious for $0\leq x\leq1$ because \begin{align} &1260-840x-210x^2+126x^3-84x^4+60x^5-45x^6+37x^7-280x^8 \\ =&840(1-x)+210(1-x^2)+210(1-x^8)+84x^3(1-x)+45x^5(1-x)+42x^3(1-x^5) \\ &+15x^5(1-x^3)+13x^7(1-x)+24x^7 \\ \geq&0. \end{align} Done!

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    @ Michael Rozenberg sorry for mistakes, I just made the corrections – 2017-02-25
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    I feel you should include a few more in-between steps in your answer – 2017-02-25
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    @mrnovice What is your question? I am ready to answer. – 2017-02-25
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    @MichaelRozenberg `obvious for 0≤x≤1` Obviousness aside, but why $x \le 1$? OP's question makes no mention of range restrictions. – 2017-02-25
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    @dxiv The topic-starter changed the given many times. The last version was with $0\leq x\leq1$ and I missed to edit it. I am sorry. – 2017-02-25
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    @MichaelRozenberg The [edit history](http://math.stackexchange.com/posts/2160339/revisions) does not show the OP mentioning $0 \le x \le 1$ at any point. In fact, this condition only appears in *your* very latest edit of the title. (Note that I am not saying that the condition doesn't makes sense, but only that it was never mentioned by the OP to begin with.) Edits that materially *change* the question are generally frowned upon. – 2017-02-25
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    @dxiv I saw how Rahul edited his post. Ask him. – 2017-02-25
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    Why is the last step true? – 2017-02-26
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    @Hans Because $840\geq840x$, $210\geq210x^2$, $84x^3\geq84x^4$, $45x^5\geq45x^6$, $37x^7\geq37x^8$ and $210+42x^3+15x^5-243x^8\geq0$. – 2017-02-26
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    Thank you! But could you please further explain the last inequality? From its plot, the root closest to $1$ is about $1.013$, which is very close to $1$, so I suppose the estimate should be very tight. – 2017-02-26
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    @Hans Because $210\geq210x^8$ and $33x^3\geq33x^8$. – 2017-02-26
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    Ah, yes. Very clever! And I have been obtuse. Would you mind writing it out in your answer of posterity so to speak --- even though it is too obvious for you? Thank you. – 2017-02-26
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    I am wondering where the upper bound comes from. Could it be a Pade approximant or a continued fraction approximation? – 2017-02-27
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$\ln(1+x) \le x-\cfrac{x^{3}}{2(1+x)}\;$ fails at $x=2$ for example, since $\ln(3) \gt 1 \gt \cfrac{2}{3} = 2 - \cfrac{2^3}{2\cdot 3}\,$.

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    The topic-starter changed the given many times. The last version was with $0\leq x\leq1$ and I missed to edit it. I am sorry. – 2017-02-25
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    To later readers, please note that the condition $0 \le x \le 1$ was inserted into the title *after* my answer was posted. – 2017-02-25
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    I saw how Rahul edited his post. Ask him. – 2017-02-25