Let $f$ be a twice derivable function and $M_i =\sup_{x \in \mathbb{R}} |f^{(i)}(x)|$ and $|M_0|, |M_2|<\infty $. Preferably using the Taylor series on the interval $[x,x+h]$ show the following inequalities: (a) $$|f'(x)| \le \frac{2M_0}{h} +\frac{hM_2}{2}$$ and (b) $$M_1 \le 2\sqrt{M_0M_2}$$ Any hints and/or a solution would be appreciated. P.S. I know that $(b) => (a)$ […]

How do I show that: $$\log(2)=\sum^\infty_{n=1}(-1)^{n+1}\frac{1}{n}$$ and that $$\log(5)=\log(3)+\sum^\infty_{n=1}(-1)^{n+1}\frac{2^n}{n3^n}$$ Thanks in advanced.

How can I prove that $$\displaystyle x-\frac {x^2} 2 < \ln(1+x)$$ for any $x>0$ I think it’s somehow related to Taylor expansion of natural logarithm, when: $$\displaystyle \ln(1+x)=\color{red}{x-\frac {x^2}2}+\frac {x^3}3 -\cdots$$ Can you please show me how? Thanks.

I am trying to focus on the limits of functions with similar series expansions and I stumbled on this. $$\lim_{x\to\infty}\left({\left(\frac{x^2+5}{x+5}\right)}^{1/2}\sin{\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)}-(x-5)^{1/2}\sin{\left({\left(x^2-5x+25\right)}^{1/2}\right)}\right)=0$$ I heard the mean value is possible but the entire function is not bounded. I can take the taylor series at infinity however the terms would be undefined. I could use substitution with the taylor […]

My professor made this claim about Taylor Series convergence in my Complex Variables class and I am still not entirely convinced (he said it’s explained in the textbook and textbook states, “we will soon see that it is impossible for the series to converge to $f(z)$ outside this circle” but they don’t actually show that). […]

I would like to show that : $$(-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$ My proof: Note that : \begin{align*} e^{x}&=1+x+\mathcal{O}\left(x^{2}\right)\quad (x\to 0)\\ \tan(x)&=x+\mathcal{O}\left(x^{3} \right)\\ \tan(a+b)&=\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\\ x^{a}&=e^{a\ln(x)}\quad (x>0) \end{align*} then: $-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)=\left(1-\tan^{-1}\left(\dfrac{1}{n}\right) \right)^{-1}$ $\dfrac{1}{n} \underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow }0$ $\left(\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^{3}} \right)\right)^{-1}=\left(\dfrac{1}{n}\right)^{-1}\left(1+\mathcal{O}\left( \dfrac{1}{n^{2}}\right) \right)^{-1}=n\left( 1-\mathcal{O}\left( \dfrac{1}{n^{2}}\right)\right)$ Thus : \begin{align*} (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}&=(-1)^{n}e^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)\ln(n)} \\ &=(-1)^{n}\exp\left[-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} […]

The proof for this $$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$ using the MacLaurin series is all right for a high school level, but I dont understand why the series that has been derived for the reals should hold for complex numbers too. Could someone give a sufficient reason why it is correct to use […]

Let $T_\epsilon=e^{i \mathbf{\epsilon} P/ \hbar}$ an operator. Show that $T_\epsilon\Psi(\mathbf r)=\Psi(\mathbf r + \mathbf \epsilon)$. Where $P=-i\hbar \nabla$. Here’s what I’ve gotten: $$T_\epsilon\Psi(\mathbf r)= e^{i \mathbf{\epsilon} P/ \hbar}\Psi(\mathbf r)=\sum^\infty_{n=0} \frac{(i\epsilon \cdot (-i\hbar \nabla)/\hbar)^n}{n!} \Psi(\mathbf r)=\sum^\infty_{n=0} \frac{(\mathbf \epsilon \cdot \nabla)^n}{n!}\Psi(\mathbf r)= \Psi(\mathbf r) + (\epsilon \cdot \nabla) \Psi(\mathbf r) + \frac{(\epsilon \cdot \nabla)^2 \Psi(\mathbf r)}{2} + […]

Use Taylor’s Theorem to estimate the error in approximating $\sinh 2x$ by $2x + 4/3x^3$ on the interval $[-0.5,0.5]$. For this question, I use the Taylor’s remainder formular, $$ R_n(x)= \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!}$$ and I get $R_4(0.5) = 0.012\,85$. Is this correct?

Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y – \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series. thanks…

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