# Tag Info

7

The rule that you quote is only valid if a is a constant.

6

$A^2=-I$ does not imply that $A=iI$. For example, one such $A$ is $$\left(\begin{array}{rr} 0 & -1 \\ 1 & 0\end{array}\right).$$ Now, for its exponential, note first that $$A^{2n}=(A^2)^n=(-I)^n=(-1)^nI$$ and $$A^{2n+1}=(A^2)^nA=(-I)^nA=(-1)^nA.$$ Then $$\exp(t A)=\sum_{n=0}^\infty\frac{t^n}{n!}A^n= ... 5 I will assume that this is not homework and go ahead to give a full solution. As others have said, your rule only works for constant a. Following the hint in danielson's answer and setting y=f(x), you have$$\log y=\sin x\log\left(1+x^2\right)$$and differentiating both sides with respect to x and using the chain and product rules gives ... 5 The comments that suggest f(x)=1 and f(x)=0 as solutions are correct, but these are just special cases of f(x)=a^x with a=1 or a=0. If f is not continuous then f need not be an exponential function on the irrationals, but on \mathbb{Q} f(x)=a^x if it satisfies this property. To see this, let f(1)=a and note that f(n+1)=f(n)f(1) so that ... 3 This, of course, uses three interconnected formulas: e^{ix}= cos(x)+ i sin(x), cos(x)= \frac{e^{ix}+ e^{-ix}}{2}, and sin(x)= \frac{e^{ix}- e^{-ix}}{2} Your error is that you are assuming that the imaginary part of e^{ix} is "i sin(x)". That is true only if itself is real. If x is not real the i sin(x) is not imaginary because sin(x) is not ... 3 Start with$$f(x)=e^{\ln(x^2+1)\sin(x)}$$The chain rule gives$$f'(x)=e^{\ln(x^2+1)\sin(x)}\times (\ln(x^2+1)\sin(x))'=(x^2+1)^{\sin(x)}(\ln(x^2+1)\sin(x))'$$Take it from here. 3 As you already said, the inequality |e^{i\phi}-1|\leq|\phi| is clear from geometry. It can for example be derived as$$ |e^{i\phi}-1|^2 = 2 - 2 \cos \phi = 2 - 2 \cos 2 \frac\phi 2 = 2 - 2 \cos^2 \frac\phi 2 + 2 \sin^2 \frac\phi 2 = 4 \sin^2 \frac\phi 2 \le 4 \big(\frac\phi 2 \bigr)^2 = \phi^2 \, . $$For a, b > 0 and 0 \le \gamma \le 1 you ... 2 The rule that you have given is for \textit{constant} a, while the base of your function is a \textit{function}. So the rule does not apply here. To fix this, let f(x) = (x^2 + 1)^{\sin(x)} and consider \ln(f(x)). Differentiate the new function, while also utilizing a nice property of exponents inside logarithmic functions, and try to deduce the ... 2 An easy induction shows that$$ A^n=\begin{cases}\hphantom{-}I&\text{if}\enspace n\equiv 0\mod4\\ \hphantom{-}A&\text{if}\enspace n\equiv 1\mod4\\ -I&\text{if}\enspace n\equiv 2\mod4\\ -A&\text{if}\enspace n\equiv 3\mod4 \end{cases} There results that \begin{align*} \exp (\varphi A)&=\sum_{k=0}^\infty \frac{\varphi^{2k}}{2k!} ... 2 It depends on the properties you want the function to have. At least for x\in\mathbb Z you are right, as x^x is well-defined for x\in\mathbb Z,x<0. When talking about x\in\mathbb Q, things get more difficult and I wouldn't argue that you can (easily) calculate f(x) with x=-\frac{2}{5}. If you choose to include all x\in\mathbb Z with ... 2 You even have -x < \operatorname{ln}(b), $$because ln \colon (0, \infty) \to \mathbb R is strictly increasing on its entire domain. 2 Write:$$ (e^x+1)^{1/x}=(e^x)^{1/x} (1+e^{-x})^{1/x}=e\times (1+e^{-x})^{1/x} $$which goes to e as x \to \infty 2 Apply the squeeze theorem:$$\lim_{x\to\infty}(e^x)^{\frac1x}\leq \lim_{x\to\infty}(e^x+1)^{\frac1x}\leq \lim_{x\to\infty}(e^x\cdot e)^{\frac1x}$$1 The condition f(x+y)=f(x)f(y) only implies f(x)=a^x for all rational numbers x\in\mathbb{Q} and for some a\in\mathbb{R}. You can get this equality for all real numbers if you have more conditions, for example, if f is continuous in \mathbb{R} or if f is Lebesgue-measurable. 1 HINT: Using the equation f(x+y)=f(x)f(y), If f(x) is polynomial, show that the degree of L.H.S. \not = degree of R.H.S. (except if f(x) is the zero polynomial) If f(x) is trigonometric, show that it is not possible using the expansion formulae of \sin,\cos,\tan of (x+y). 1 Well, I suppose you could leave e^z undefined for complex z. The fact that \exp(z) agrees with e^x on the real line, and is the only so-called "analytic" function which agrees with e^x on the real line, is the reason we define e^z=\exp(z). For example,$$\exp(z)=\lim_{n\to\infty}\left(1+\frac{z}{n}\right)^n$$But it really is worth thinking of ... 1 First, "\exp(x)" is notation. It is defined to be identical to \mathrm{e}^x, whatever that is. Second, there is nothing challenging about using complex exponents: a^{x+ \mathrm{i} y} = a^x a^{\mathrm{i}y}. However, to simplify further, we need to know what to do with those imaginary exponents. When you study calculus of a real variable, you ... 1 Hint: 1 Both sides satisfy the same differential equation and have the same initial conditions, so they agree. More explicitly, let f(x)=e^{iAx} and let g(x)=I\cos x + i A \sin x. These are matrix-valued functions of x. Compute the following:$$f'=iAe^{iAx},\quad f''=-e^{iAx},\quad g'=-I\sin x+i A\cos x,\quad g''=-I\cos x -iA\sin x.$$Therefore f''=-f and ... 1 As said in comments and answers, there is no explicit solution to the equation which, after integration write$$f(x)=\frac{1}{2} e^{\frac{1}{2 \sigma_1^2}} \text{erfc}\left(\frac{x}{\sqrt{2} \sigma _1}\right)+\frac{1}{2} e^{\frac{1}{2 \sigma_2^2}} \text{erfc}\left(\frac{x}{\sqrt{2} \sigma_2}\right)-a=0f'(x)=-\frac{e^{\frac{1-x^2}{2 ...

1

Following Sylvain's comment look up the formulas for Fourier transform $F(w)=\int f(t) e^{-iwt}{\rm {d}}t$ and the inverse transform $f(t)=1/{2\pi}\int F(w) e^{iwt} { \rm {d}}w$ and combine them to write \begin{split} F(\hat w)=\int_{-\infty}^\infty f(t) e^{-i\hat{w}t}\rm {d}t & = \int_{-\infty}^\infty ...

1

$$\lim_{n\to \infty} \left(\frac{n}{n-1}\right)^n=\lim_{n\to \infty} \left(1+\frac{1}{n-1}\right)^{n-1} \cdot \left(1+\frac{1}{n-1}\right)= e.$$

1

For your second insight... $$\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x}$$ You can split the limit into product terms if and only if the product terms both exist. i.e., before splitting into a product, observe that $$\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x-1} = \lim\limits_{t \to \infty}\left(\dfrac{t+1}{t}\right)^{t}$$ using ...

1

$$\left(\frac{x}{x-1}\right)^x=\left(\frac{x-1+1}{x-1}\right)^x=\left(1+\frac{1}{x-1}\right)^x=\left[\left(1+\frac{1}{x-1}\right)^{x-1}\right]^{\frac{x}{x-1}}$$ When $x\to \infty$, the exponent $\frac{x}{x-1}=\frac{1}{1-1/x}$ tends to $1$ and $\left(1+\frac{1}{x-1}\right)^{x-1}$ it is known tends to $e$.

1

If you know that $$\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{x}=e^a$$ then $$\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^{x}=e^{-1}$$ and so $$\lim_{x\to\infty}\left(\frac{x}{1-x}\right)^x= \lim_{x\to\infty}\frac{1}{\left(\dfrac{x-1}{x}\right)^x}= \lim_{x\to\infty}\frac{1}{\left(1-\dfrac{1}{x}\right)^x}= (e^{-1})^{-1}=e$$ The proof of the first ...

1

I believe there are tons of such functions. One of them is: $$e^x \cos (2\pi x)$$

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