Tag Info

Hot answers tagged

24

Every non-zero complex number $z$ has exactly two complex square roots - this is a consequence of the field of complex numbers being algebraically closed (Wikipedia link). If $$z=re^{i\theta}=r\cos(\theta)+ri\sin(\theta)$$ then the square roots of $z$ are $$\begin{align*} \sqrt{r}e^{i\theta/2}&=\sqrt{r}\cos(\theta/2)+\sqrt{r}\,i\sin(\theta/2)\\ ...


11

Maybe you want to solve $z^2=i$. It is easy to verify that $z=\pm\frac{\sqrt2}{2}(1+i)$ satisfy the equation. Generally, every polynomial with complex coefficient has a root in $\mathbb C$, or, equivalently, complex field has no algebraic field extension.


11

The following (not particularly elegant) proof uses reasonably basic multiplication and division. We need to show that $7^{31} > 8^{29}$, i.e. that $\dfrac{7^{31}}{8^{29}}>1$. We have: ...


9

HINT: Divide by $4^x$ to get $$a^2-a-2=0$$ where $a=\left(\dfrac32\right)^x$ Can you solve for $a?$ Now for real $x,a>0$ See also : Exponent Combination Laws


8

The square root function is not as nicely behaved on the complex numbers, $\Bbb C$ as it is on the (nonnegative) real numbers, $[0, \infty)$. It's true that we can find exactly two solutions to the equation $w^2 = z$ for any nonzero $z \in \Bbb C$, but unlike in the usual real setting, we cannot make choice of $w$ that continuously depends on $z$, or more ...


7

Take logs, to get $b\ln a>a\ln b$, or $\frac{\ln a}a>\frac{\ln b}b$. The function $\frac{\ln x}x$ increases to a maximum at $x=e=2.71828...$, then decreases. So your question is true if $e<a<b$ If you don't like logs, take the $ab^{th}$ root of both sides, and get $\sqrt[a]a>\sqrt[b]b$ when $e<a<b$.


7

so start with for $a>1$ and $\forall n$ big enough we have $$n^a < a^n \Leftrightarrow \log(n^a)<\log(a^n)\Leftrightarrow a\log(n)<n\log(a)\Leftrightarrow 1<\frac{\log(a)}{a}\frac{n}{\log(n)} $$ and we fixed $a$, so $\frac{\log(a)}{a}=c>0$ is just a constant. So in fact we have to show, that $$ 1<c\frac{n}{\log(n)} $$ holds for all ...


7

One should not say "equals undefined"; one should say "is undefined". The is the "is" of predication, not the "is" of equality. $0^0$ is indeterminate in the sense that if $f(x)$ and $g(x)$ both approach $0$ as $x\to a$ then $f(x)^{g(x)}$ could approach any positive number or $0$ or $\infty$ depending on which functions $f$ and $g$ are. But in some ...


7

Let $r$ be the random number, and let $r_n$ be the number obtained after repeating the 4 step procedure $n$ times. Let $b$ be the base used in steps 1 and 3. Starting with $r_n$, we do the following: 1 Exponentiate: $b^{r_n}$ 2 Raise to the power $x$: $(b^{r_n})^x = b^{xr_n}$ 3 Take the log (same base): $\log_b(b^{xr_n}) = xr_n$ 4 Take the $x$-th root: ...


6

Others may bristle at this "proof," but: $$7^{31} = 157,775,382,034,845,806,615,042,743 \\ 8^{29} = 154,742,504,910,672,534,362,390,528$$ If all else fails, just calculating the expressions and comparing them will work. This particular problem is only mildly tedious to attack this way if you have pen/paper.


6

You can rewrite this as $$(e^x)^2-6-e^x=0$$ $$(e^x+2)(e^x-3)=0$$ I'll let you take it from here


6

HINT : Multiply the both sides by $e^x$ to get $$(e^x)^2-6-e^x=0.$$ Now, let $e^x=t$.


5

let $e^x = a$ $$ a - \frac{6}{a} - 1 = 0$$ $$ a^2 - a-6 = 0$$ $$ a = 3 \ or \ -2 $$ $$ e^x = 3$$ Edit: for $x \in \mathbb{R}$ Can you find the value of x now? Hint: take $\ln$ of both sides.


5

It's the same trick, except easier. Pull out all the $2$'s and $5$'s to see how many $10$'s you get: $$ 15^{80}28^{60}55^{70} = 5^{80}4^{60}5^{70}(3^{80}7^{60}11^{70}) = 2^{120}5^{150}(3^{80}7^{60}11^{70}) = 10^{120}(5^{30}3^{80}7^{60}11^{70}) $$


5

To clarify one thing (Michael already posted a great answer). You don't prove something is undefined. Undefined is not some mystical object. Undefined means we have not defined it. You can prove something can't be defined in a way consistent with other math. I can define $0^0$ anyway I like. I define $0^0=54$. There, now it is defined, but it is not ...


4

HINT: multiply both sides by $e^x$ to turn it into a quadratic equation in $e^x$.


4

Basically instead of degrees and radians, on a complex plane we can use multiplication to express rotations. Multiplying by $1$ takes you take to the same place, so it's the equivalent of 360 degrees. Multiplying by $-1$ is the same as 180 degrees. Multiplying by $i$ is the same as a 90 degree rotation and multiplying by $\sqrt{i}$ is the same as rotating ...


4

The more elegant way to solve this equation was given by lab bhattacharjee, but I'll simply elaborate on how to solve it using your method, which was still correct. Hint: We have that $$a^2 - ab -2b^2 = 0 \iff b = \frac{a}{2}, \text{or }b= -a.$$ You now have a system of two equations, $\log_3 a = \log_2 b$ and from this you can see that neither $a$ or ...


4

This is what happens when you apply finite differences to any sequence. Here is some useful notation. If $a_n$ is a sequence, its forward difference is the sequence $$\Delta a_n = a_{n+1} - a_n.$$ (The notation should not be read as "$\Delta$ of $a_n$," but as "the $n^{th}$ term of the sequence $\Delta a$.") For example, if $a_n = n^2$, then $$\Delta a_n ...


3

Nice observation -- you have found one of the fundamental properties of finite differences. Let me rewrite your question in a more "functional" way: Fix an abelian group $A$, written additively. (For instance, $A$ can be $\mathbb{Z}$ or $\mathbb{Q}$ or $\mathbb{C}$. Either choice works for your question, so if you are not on friendly terms with groups, you ...


3

Try to make a variable change of type: $$ g(n) = f(n) + n^{\alpha} a $$. Try $\alpha$ from $0, 1, 2 ..$. Here is the case where $\alpha = 0, 1$ does not work. But $\alpha = 2, a = -\frac{c}{2}$ works! Then $g(n)$ satisfies: $$ g(n) = 2g(n-1) - g(n-2) $$


3

$$64/100 = .64$$ So then you have $.8^d = .64$ What does $.8 \times .8$ equal?


3

Let $f(x) = e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}-1 =\sum_{i=1}^N e^{k_i/x}-1 $, and let $K = \sum_{i=1}^N k_i $. The restrictions that $x > 0$ and $k_i < 0$ are important in what follows. $f'(x) =\sum -\frac{k_i}{x^2}e^{k_i/x} =-\frac1{x^2}\sum k_ie^{k_i/x} =\frac1{x^2}\sum |k_i|e^{k_i/x} $, so $f'(x) > 0$. This means that your function has at ...


3

Consider $$A(\theta) = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0 & \cdots & 0\\ \sin(\theta) & \cos(\theta) & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1\end{bmatrix} $$ Then $A(\theta)^n = A(n\theta)$. ...


3

Other answers have explained why $\sqrt{i}$ is a complex number, and shown how to compute it using complex exponentials, but you can also compute it directly. If you want to find the complex number $a + bi$, where $a$ and $b$ are real numbers, such that $\sqrt{i} = a + bi$, then square both sides and solve $$i = (a + bi)^2.$$ Expanding the right-hand side, ...


3

We can show that the only $nice$ solutions to $$f(x^y)=f(x)^{f(y)}\qquad(1)$$ are the identity function, and the constant functions $1$ and $-1$. There are also a few $ugly$ solutions. To avoid problems with $0^0$, let's first find all functions $f$ such that $f:(0,+\infty)\to(-\infty,0)\cup(0,+\infty)$. (Note that because the question is about nonnegative ...


3

Your last step doesn't work, instead do: $$ 3^{3x - (4x - 2)} = 3^{x+4}. $$


3

I would follow the pseudocode given here: Write your exponent, 1357, in binary: $10101001101_2$. Let $b := x\mod 2623$. Let $r := 1$. Step through the bits from right-to left: If the bit is $1$: Let $r := b \cdot r \mod 2623$. Let $b := b^2 \mod 2623$. Then $r$ will be your final result. This requires 17 multiplications modulo $2623$. In general it ...


3

It is no longer true that $$\sqrt{ab}=\sqrt{a}\sqrt{b}$$ when $a,b$ are not positive real numbers. Additional remark: One must be careful when talking of $\sqrt{\cdot}$ in the realm of complex numbers, because it cannot be defined globally, i.e. there is no continuous function $$f:\Bbb C\to \Bbb C.$$ such that $f(z)^2=z$ for all $z\in\mathbb{C}$. Formally, ...


3

If we know (with suitable restrictions on $k$ and $n$) that $k^2 = n$, then we can say $$ k = \sqrt{n} = n^{1/2}$$ So simply cubing both sides yields the desired $$\bbox[10px, border: solid blue 1px]{k^3 = (\sqrt{n})^3 = \sqrt{n^3}= n^{3/2}}$$ So your original answer was correct. $k$ is indeed equal to $n^{1.5}$, your error is probably found elsewhere.



Only top voted, non community-wiki answers of a minimum length are eligible