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15

HINT: Using Exponent Laws(1) , (2) $$n^{a+b}=n^a\cdot n^b$$ Setting $a=0,$ $$n^b=n^b\cdot n^0$$ Check when $n^b$ can be cancelled


7

You can use Newtons approximation method. We want to find $x=2^{1.4}$, or equivalently, $x^5=\left(2^{1.4}\right)^5=2^7=128$ Define $f(x)=x^5-128$ We want to find the root of $f(x)$ as noted in my comment, $x\approx \sqrt{8} \approx 2.828$ So we start with this guess of $2.828$. We put it into the formula, $$x'=x-\frac{f(x)}{f'(x)}$$ and $x'$ will ...


4

\begin{align*} n^m &= n\times \ldots \times n\; (\textrm{for}\; m \in \mathbb{Z}_{>0})\\ &\vdots\\ n^2 &= n\times n\\ n^1 &= n\\ n^0 &= 1\\ n^{-1} &= \frac{1}{n}\\ n^{-2} &= \frac{1}{n\times n}\\ &\vdots\end{align*} To go from $n^{m}$ to $n^{m+1}$, you multiply by $n$, and to go from $n^{m}$ to $n^{m-1}$, you divide by ...


4

When you do $m\cdot n$, you start from $0$ and keep adding $n$ as many times as $m$ says: so if $m=0$, you just have $0$ because you have nothing else to do; if $m=1$, you get $0+n$; if $m=2$ you do $0+n+n$. And so on. Exponentiation is exactly the same, but with multiplication replacing addition and $1$ replacing $0$: for doing $n^m$ you start from $1$ and ...


4

This is probably not a complete answer but still worth knowing. In general, to compute $a^n$, you do not need to multiply it $n-1$ times. You can get away with at-most $\mathcal{O}(\log_2(n))$ multiplications as shown below. Write $1038$ in binary as $2^{10} + 2^3 + 2^2 + 2$. Now to compute $a^{1038}$, it is now sufficient to compute only $a^2$, $a^4$, ...


3

HINT: $$a^x\cdot b^x= (a\cdot b)^x$$ for positive real $a,b$ and real $x$ but $$a\cdot b^x\ne (a\cdot b)^x$$ in general In fact if finite $a\cdot b\ne0$, $$a\cdot b^x= (a\cdot b)^x\implies a^{x-1}=1$$ $\displaystyle\implies $ either $a=1,$ or $x-1=0$ or $a=-1,x-1$ is even


3

Algebraically, substituting $x = \pm\pi$ into Euler's Formula, $e^{ix} = \cos x + i \sin x$ yields: $$\begin{align}e^{\pm i\pi} &= \cos(\pm\pi) + i \sin (\pm\pi) \\ &= -1 + 0\cdot i\\ &= -1\end{align}$$ Geometrically, $e^{\pm i\pi}$ is an anti-clockwise/clockwise rotation of the point corresponding to $1$ on the Argand diagram by $\pi$ radians, ...


3

The first thing to note is that you have the derivative with respect to $x$, but $x$ is not in your equation. Technically: $\frac{d}{dx}e^{4y} = 0$ You should write: $\frac{d}{dy}e^{4y}$ Then we can get that with the chain rule. Let's go through the steps explicitly. $ F = e^{4y}$ to: $ F = u^{y} $ by saying that $u = e^{4}$. The chain rule would ...


3

I figured out the problem: The solution to $\color{blue}{\sqrt 3}(x) = x + \sqrt 3$ is indeed $$\frac {1}{2} \sqrt{3} + \frac {3}{2}$$ But that's not the problem you posted. In the above, only $3$ is under the radical sign. In your post, you have $\sqrt{3x}$ In the event that the problem should read: $$\sqrt 3(x) = x + \sqrt 3$$ then $$\begin{align} \sqrt ...


3

I read this post a while ago, and since then I found an extremely good approximation for $e^x$ $e^x=1+\cfrac{2x}{2-x+\cfrac{x^2}{6+\cfrac{6x^2}{10+\cfrac{10x^2}{14+...}}}}$ Truncated here, this gives an error of less than $10^{-7}$ for $x\in(-1.27,1)$. Every additional row increases this range by about 0.4 on both sides, but by using the repeated squaring ...


2

$\displaystyle -y = \sqrt{(x+y)(y+z)}\implies(x+y)(y+z) = (-y)^2=y^2$ $\displaystyle\implies y^2=(x+y)(y+z)=xy+y^2+xz+yz\iff xy+yz+zx=0 $ If $\displaystyle x=0,yz=0\implies$ at least one of $y,z$ is $0$ If $y=0,$ from the given relation becomes an identity Assuming $xyz\ne0,$ divide either sides by $xyz$


2

Instead of translation of a function $f(x) \to f(x+a)$ , let us consider scaling. This means that we are going to make intervals of the independent variable smaller, or larger, with a factor $\lambda > 0$. The transformed function is then defined by: $$ f_\lambda(x) = f(\lambda\,x) $$ Like with translations, it would be nice to develop the function ...


2

In the integral formula, you have $$f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(a)}{a-z}\,da.$$ Now, with the parametrisation $\gamma(t) = z + re^{it}$ and $a\: \hat{=}\: \gamma(t)$, the differential $da$ becomes $$da = \gamma'(t)\,dt = ire^{it}\,dt = i(\gamma(t)-z)\,dt,$$ or, in other words, $$ \frac{da}{i(a-z)} \:\hat{=}\: dt.$$ The $i$ is cancelled ...


2

Proposition: Let $E=\{A\in M_n(\mathbb{C})|A\text{ has algebraic entries }\}$. Then the exponential map is injective on $E$. Proof: Assume that $e^A=e^B$. Here $A,B$ have algebraic entries ; according to "Wermuth, 2 remarks on matrix exponentials" (in free access) http://www.sciencedirect.com/science/article/pii/0024379589905545 $AB=BA$. Thus $A,B$ are ...


2

It is enough to find the minimum value of $$ f_a(x):=\cos^6x+\sin^6x+a\cos x\sin x $$ We have \begin{eqnarray} f_a(x)&=&\cos^6x+\sin^6x+a\cos x\sin x\\ &=&(\cos^2x+\sin^2x)^3-3(\cos^4x\sin^2x+\cos^2x\sin^4x)+a\cos x\sin x\\ &=&1-3\cos^2x\sin^2x(\cos^2x+\sin^2x)+a\cos x\sin x\\ &=&1-3\cos^2\sin^2x+a\cos x\sin x\\ ...


1

Let's ignore the insignificant part $2^{105}$ first $$ x = 2^{129} \\ \log_{10} x = 129 \times \log_{10}2 \approx 38.83286944 \\ \therefore x \approx 10^{38.83286944} = 10^{0.83286944} \times 10^{38} \approx 6.805647 \times 10^{38} $$ Similarly, we can get $2^{105} \approx 4.056 \times 10^{31}$.


1

Something very useful is to notice that $2^{10} = 1024 \approx 10^3$ You calculations are almost correct (wrong sign in front of 105) but you can neglect $2^{105}$ because it is very small compared to $2^{129}$ So :$$2^{129} = 2^9(2^{10})^{12}\approx 512\times (10^3)^{12} = 5.12\cdot 10^2\times10^{36}$$ This way you can find a very good estimate of you ...


1

Consider $f(x) = 3^x + 10 - 2*7^x$, and $f(0) = 9 > 0$, $f(1) = -1 < 0$, So by IVT, there is $0 < c < 1$ such that $f(c) = 0$. If $x < 0$, then $f(x) > 0$, and if $x > 1$, then $f'(x) = ln3*3^x - 2ln7*7^x < 0$, so $f(x) < f(1) = -1$. So there is no solution for $x < 0$ or $x > 1$. So you may use Newton's approximation method ...


1

I don't think there is a closed form solution to this equation. Had it be $3^x +10=2.9^x=2.3^{2x}=2.(3^x)^2$ instead, you could have set $t=3^x$ to turn it to an algebraic equation, easily solved: $t+10=2.t^2$. Trying the same trick here leads to $t+10=2.t^{\frac{log(7)}{log(3)}}$. The exponent ($1.7712437...$) is irrational so that there is no way of ...


1

The first matrix can be factored as $A=\begin{bmatrix}1& 1\\ 0& 1\end{bmatrix}=\begin{bmatrix}1& 0\\ 0& 1\end{bmatrix}+\begin{bmatrix}0& 1\\ 0& 0\end{bmatrix}=X+N$ So, $e^A=e^Xe^N=\begin{bmatrix}e& e\\ 0& e\end{bmatrix}$ The second is digonalizable: $B=\begin{bmatrix}-6& -2\\ 3& ...


1

For a Jordan Block, we have $A = I + N$: $$e^A = e^{j_\lambda} = e^{\lambda I + N} = e^{\lambda}\sum_{j=0}^{n-1} \dfrac{1}{j!}N^j$$ This gives: $$e^A = e\begin{bmatrix}1& 1\\ 0& 1\end{bmatrix} = \begin{bmatrix}e& e\\ 0& e\end{bmatrix}$$ For the second, we can diagonalize the matrix as: $$A = PJP^{-1} = \begin{bmatrix}-1& -2\\ 2& ...


1

I don't know if you're allowed to use that Log(2) = .693147... or not. If you are you can do the following: 2^1.4 = 2 2^.4 = 2 2^.5 2^-.1 . Now you need only approximate 2^-.1 . 2^-x for small x: 2^-x = 1 - Log[2] x + 1/2 Log[2]^2 x^2 - 1/6 Log[2]^3 x^3 + ... Now set x=0.1 and keep as many terms as needed for your desired accuracy. You get 2^x = 1, ...


1

Any continuous one-to-one function $f(x)$ can form the basis for a so-called quasi-arithmetic mean (aka Kolmogorov mean) by defining the average $<x>$ of the numbers $x_1, x_2, .. x_n$ as $$n f(<x>) = f(x_1) + f(x_2) + .. + f(x_n) $$ For $f(x) = x$ the arithmetic mean is recovered, for $f(x) = 1/x$ we get the harmonic mean, and for generic ...


1

You can do something like: \begin{eqnarray} \frac{3}{4t^{1/4}}-\frac{1}{2t^{3/4}} &=& \frac{3}{4t^{1/4}}\frac{t^{2/4}}{t^{2/4}} -\frac{1}{2t^{3/4}}\frac{2}{2} \\ &=&\frac{3 t^{2/4}}{4t^{3/4}} - \frac{2}{4t^{3/4}} \\ &=&\frac{3 t^{1/2}-2}{4t^{3/4}} \\ &=&\frac{3 \sqrt{t}-2}{4t^{3/4}} \end{eqnarray} The point is that you make ...


1

They are equivalent. Let $e^k = 1+r$, and the first becomes $N_0 (1+r)^t$ and the second becomes $N_0 e^{kn}$. $n$ and $t$ just measure how far things have gone, with $n$ usually having integral values and $t$ usually having continuous values. If $t$ is measured at equally spaced intervals, then $t = a+bn$, so $N_0(1+r)^t =N_0(1+r)^{a+bn} =N_0 ...



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