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$f: (-\frac{\pi}{2}, \frac{\pi}{2}) \rightarrow \mathbb{R}, f(x) = \log(\cos x)$

Count the Taylor-polynomial $T_{2}(f, 0)(x)$ of the second degree of $f$ in $x_{0} = 0$

Alright because it was saying second degree, I knew we will need one derivation of $f$ at least (is that correct?). (Is it allowed to use more derivations if the task says second degree by the way?)

$f^{0}(x) = \frac{\ln(\cos x)}{\ln(10)}$

$f^{1}(x) = -\frac{\sin x}{\cos x\cdot \ln(10)}$

Put that into the Taylor formula: $(T_{n,x_{0}}f)(x)=\sum_{k=0}^{\infty}\frac{1}{k!}f^{k}(x_{0})(x-x_{0})^{k}$

$\Rightarrow$

$$\log(\cos x) = \frac{\frac{\ln(\cos(0))}{\ln(10)}}{0!}*0(x-0)^{0} + \frac{-\frac{\sin(1)}{\cos(1)\cdot \ln(10)}}{1!}*0(x-0)^{1} + ..$$

$$\log(\cos x)=0$$

No wonder if we do it at $x_{0} = 0$ that it will all result in $0$...

But I must have done something wrong, don't think it's as easy as I did above, or is it? :o

Because now I cannot really create a series with this solution.

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  • $\begingroup$ $$\int\tan x\ dx=?$$ $\endgroup$
    – lab bhattacharjee
    Jul 25, 2016 at 11:13
  • $\begingroup$ I'm not allowed to use / work with integrals :( $\endgroup$
    – cnmesr
    Jul 25, 2016 at 11:14
  • $\begingroup$ For second degree, you also need the second derivative. Do they really want $\log_{10}$, or is $\log$ here the name of the natural logarithm? $\endgroup$
    – André Nicolas
    Jul 25, 2016 at 11:20
  • $\begingroup$ Thanks for the answer! One of my question is clear then. But I don't know if it's meant to base 10 or natural, not mentioned anywhere (on the example exam).. $\endgroup$
    – cnmesr
    Jul 25, 2016 at 11:22
  • $\begingroup$ In mathematics, the symbol "$\log{x}$" means "logarithm to a base and it should be perfectly obvious to you what the base is", while "$\ln{x}$" means "logarithm to base $e$". In a text that alternates $\log{x}$ and $\ln{x}$, $\log{x}$ often means $\log_{10}{x}$. In a text that involves calculus and doesn't mention "$\ln{x}$", "$\log{x}$" usually means "$\log_e{x}$". In a computer science text that doesn't mention "$\ln{x}$", "$\log{x}$" can easily mean "$\log_2{x}$". So in this case, make life easier for yourself by assuming base $e$ rather than base $10$. $\endgroup$
    – Martin Kochanski
    Jul 25, 2016 at 11:38

2 Answers 2

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We may compute the whole Taylor series from the Weierstrass product: $$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) \tag{1}$$ and the Taylor series of $\log(1-x)$, leading to: $$ \log\cos(x) = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m\pi^{2m}(2n+1)^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\,\zeta(2m)}{m\,\pi^{2m}}\,x^{2m}.\tag{2}$$ Since $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$, in a neighbourhood of the origin we have: $$ \log\cos(x) = -\frac{x^2}{2}-\frac{x^4}{12}+O(x^6).\tag{3}$$

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  • $\begingroup$ Thank you for sharing another way of doing it! However I cannot understand it sadly because Weierstrauss theorem hard for me to understand :P $\endgroup$
    – cnmesr
    Jul 25, 2016 at 17:29
  • $\begingroup$ @cnmesr: then just make a leap of faith: $(1)$ holds for any $x\in\mathbb{C}$, and the convergence of the infinite product (together with its derivatives) is uniform over any compact subset of $\mathbb{C}$. $\endgroup$
    – Jack D'Aurizio
    Jul 25, 2016 at 17:31
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For the derivatives one get:

$$f'(x_0)=-\tan(x)$$ $$f''(x_0)=-\frac{1}{\cos^2(x)}$$ (Use the chainrule with $u=\log(x), v=\cos(x)$)

Then you get: $$T_{f,0,2}(x)=\sum\limits_{j=1}^2 \frac{f^{(j)}(0)}{n!}\cdot x^n$$ $$T_{f,0,2}(x)=0-\tan(0)(x-0)-\frac{1}{2}x^2$$ $$T_{f,0,2}(x)=-\frac{1}{2}x^2$$

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  • $\begingroup$ Thank you! Does that mean my first derivation is wrong? Oh yeah it is in deed, thank you very much again! :) $\endgroup$
    – cnmesr
    Jul 25, 2016 at 11:28
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    $\begingroup$ You're welcome. No, not at all. I'm not sure, why you used $ln(10)$ rather where it is coming from but: $\frac{-\sin(x)}{\cos(x)}=-\tan(x)$, so parts are correct. $\endgroup$
    – jacmeird
    Jul 25, 2016 at 11:31
  • $\begingroup$ I have used it because I didn't know how to derivate log(x). But I knew $log(x) = \frac{ln(x)}{ln(10)}$ Now I found my mistake I think. Derivated wrong because I have ignored the factor $\frac{1}{ln(10)}$ completely ^.^ $\endgroup$
    – cnmesr
    Jul 25, 2016 at 11:34
  • $\begingroup$ Ah, I see. It's easier to use the chainrule for the first derivative. The second derivative you can get with the quotient rule ;) $\endgroup$
    – jacmeird
    Jul 25, 2016 at 11:35

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