33
votes
Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?
This is not a new method. It's essentially:
guess
if the guess was too small, guess one bigger (or vice versa)
repeat
This is just a bit better than blind trial-and-error, and would have been ...
12
votes
Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?
This is not a new method. It actually comes from:
$$(a-b)(a+b)=a^2-b^2$$
For consecutive perfect squares, it is $a-b=1$.
For instance,
$$196 - 169 = (14-13)(14+13)$$
$$196 - 169 = (14+13)$$
$$196 = ...
5
votes
Tedious rational integral
There might be some things to attempt before substitution. First, we can factor some things out of the radicals.
$$\int\dfrac{\sqrt{x^2+x+2-\sqrt{4x^2+4x+4}}}{x\sqrt{x^4+x^3+x^2}}=\int\dfrac{\sqrt{x^...
- 13.1k
5
votes
evaluate the limit when x goes to infinity
Let $t=\frac{1}{x}$. Then
$$(x^3+6x^2+1)^{\frac13}-(x^2+x+1)^{\frac12}
=
\frac{(t^3+6t+1)^{\frac13}-(t^2+t+1)^{\frac12}}{t} = \frac{f(t)-f(0)}{t-0}
$$
where $f(t)=(t^3+6t+1)^{\frac13}-(t^2+t+1)^{\...
- 11.2k
4
votes
Accepted
Getting rid of cube roots in the form of (a+b)+(a-b)
Since $(a+b)^3=a^3+b^3 + 3ab(a+b)$
$(\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}})^3=18+5\sqrt{13} + 18-5\sqrt{13} +3\times \sqrt[3]{18+5\sqrt{13}}\times \sqrt[3]{18-5\sqrt{13}}(\sqrt[3]{18+5\sqrt{...
- 7,033
3
votes
Accepted
How to prove $\sqrt{2a+3bc}+\sqrt{2b+3ca}+\sqrt{2c+3ab}\ge 3\sqrt{5}$?
Proof.
The desired inequality is written as
$$\sqrt{\frac{2a + 3bc}{5}} + \sqrt{\frac{2b + 3ca}{5}} + \sqrt{\frac{2c + 3ab}{5}} \ge 3. \tag{1}$$
By AM-GM, we have
$$\sqrt{\frac{2a + 3bc}{5}} = \frac{...
- 36.8k
3
votes
Accepted
Find the best constant $k$ such that $\frac{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}{\sqrt{a+b+c+k\cdot abc}}\le 1+\sqrt{2}$
Some thoughts.
Remark: It is similar to my answer here.
It suffices to prove that
$$(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2 \le (3 + 2\sqrt 2)(a + b + c + k_0abc). \tag{1}$$
By Cauchy-Bunyakovsky-Schwarz ...
- 36.8k
3
votes
Accepted
Prove $\frac{1}{\sqrt{a+bc}}+\frac{1}{\sqrt{b+ca}}+\frac{1}{\sqrt{c+ab}}\ge \frac{2\sqrt{2}+1}{2},$ when $a+b+c+abc=4.$
Sketch of a proof.
Remarks: Naturally, I deal with it using AM-GM and C-S. (Fortunately, it is true.)
WLOG, assume that $a \ge b \ge c$.
By AM-GM, it suffices to prove that
$$\frac{2\sqrt 2}{a + bc + ...
- 36.8k
2
votes
Accepted
Is it possible to turn $\sqrt\frac12$ into $\sqrt2$? $ \frac{3}{4}\sqrt{\frac{1}{2} } = \frac{3}{8}\sqrt{{2} }$: What is between the left/right sides?
Well, since
$\sqrt{\dfrac{1}{2}} = \dfrac{\sqrt{1}}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}$,
it follows that
$\dfrac{3}{4}\sqrt{\dfrac{1}{2}} = \dfrac{3}{4} \cdot \dfrac{1}{\sqrt{2}}$.
We know that $\dfrac{\...
- 96
2
votes
Solving a complex ODE with complex exponents
$$(y’)^{n-1}=ny^{n-1} \iff\left(\frac{y’}y\right)^{n-1}=n$$
Remember the complex logarithm, $k\in\Bbb Z$:
$$(n-1)\ln\left(\frac{y’}y\right)=\ln(n)+2\pi i k\\\ln\left(\frac{y’}y\right)=\frac{\ln(n)}{n-...
- 10.6k
2
votes
How to integrate $\int \frac{1}{\sqrt{x+1}-\sqrt{x+3}+\sqrt{x+5}}dx$?
SOLUTION OUTLINE :
Let middle term be $\sqrt{x+b}=y$
We get $x=y^2-b$
Substitute that in the other terms to get $\sqrt{y^2+a-b}$ & $\sqrt{y^2+c-b}$
We can Partially rationalize Denominator : ...
- 7,455
2
votes
Finding $\small{\min\limits_{ab+bc+ca=1}\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}- \sqrt{2-abc}.}$
Another way.
For $k=\sqrt{\frac{8}{3}}-1$ by Holder we obtain:
$$\left(\sum_{cyc}\sqrt{a+2}\right)^2\sum_{cyc}(a+2)^2(ka+b+c)^3\geq\left(\sum_{cyc}(a+2)(ka+b+c)\right)^3$$ and it's enough to prove ...
- 197k
2
votes
Accepted
Finding $\small{\min\limits_{ab+bc+ca=1}\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}- \sqrt{2-abc}.}$
Proof.
We need to prove that
$$\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\ge \sqrt{2-abc}+2\sqrt{3}.\tag{1}$$
WLOG, assume that $a \ge b \ge c$. We have
$$\frac13 \le ab \le 1, \quad c \le \frac{1}{\sqrt{3}}. \...
- 36.8k
2
votes
Finding $\small{\min\limits_{ab+bc+ca=1}\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}- \sqrt{2-abc}.}$
Here is another approach. We assume $a,b,c$ are nonnegative.
Let us denote the constraint set as $$C=\{(a,b,c): ab\leq 1, a\geq 0,b\geq 0, c=\frac{1-ab}{a+b} \}.$$
Note that $-\sqrt{2-abc}\geq -\sqrt{...
- 56
2
votes
Finding $\small{\min\limits_{ab+bc+ca=1}\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}- \sqrt{2-abc}.}$
By your work $$\tfrac{1}{2\sqrt{a+2}}+\tfrac{bc}{2\sqrt{2-abc}}+\lambda(b+c)=\tfrac{1}{2\sqrt{b+2}}+\tfrac{ac}{2\sqrt{2-abc}}+\lambda(a+c)=\tfrac{1}{2\sqrt{c+2}}+\tfrac{ba}{2\sqrt{2-abc}}+\lambda(b+a)=...
- 197k
2
votes
Accepted
How does $\lambda \sqrt{\frac{1+v/c}{1-v/c}}$ become $\lambda \frac{1+v/c}{\sqrt{1-v^2/c^2}}$?
In such questions with a + in numerator and - in denominator or vice versa, we can multiply by $\sqrt{1+\frac{v}{c}}$ in the numerator and denominator to get rid of one of the square roots that ...
- 537
2
votes
Accepted
How to express $y$ from $x^2+y^2>r$?
First of all, the logical connective is $\vee$ and not $\wedge$; notice that there exists no real number $y$ such that $y>\sqrt{1-x^2}$ and $y<-\sqrt{1-x^2}$, so that inequality describes the ...
- 2,142
2
votes
evaluate the limit when x goes to infinity
$$\lim_{x \to \infty} (x^3+6x^2+1)^{\frac13}-(x^2+x+1)^{\frac12}$$
Using $(1+t)^a\sim1+at $ as $t\to0$, you have:
$$\sqrt[3]{x^3+6x^2+1}-\sqrt{x^2+x+1}=x\sqrt[3]{1+6/x+1/x^3}-x\sqrt{1+1/x+1/x^2}\sim \\...
- 4,252
2
votes
Getting rid of cube roots in the form of (a+b)+(a-b)
Analytically compute $\displaystyle ~\sqrt[3]{18 + 5\sqrt{13}} + \sqrt[3]{18 - 5\sqrt{13}}.$
$\underline{\text{Preliminary Considerations}}$
$$\left[ ~r + s\sqrt{13} ~\right]^3 = \left[r^3 + (39 rs^2)...
- 33.7k
2
votes
Finding the smallest root of $a(1-a)^{n-1} = s$
I think that we have a good approximation using the $[2,2]$ Padé approximant built around $a=0$
$$f_n(a)=a(1-a)^{n-1}\sim \frac {6 a-2 (n+1)a^2 } { 6+4 (n-2)a+ (n-2) (n-1)a^2}=g_n(a)$$
whose error ...
- 256k
2
votes
Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?
The proof for this method can be explained simply via telescopic cancellation method.
Consider a number $n^2=N$ whose square root $n$ we want to determine, then:
\begin{align*}
n^2 - (n-1)^2 &= n ...
- 1,273
2
votes
Tedious rational integral
Starting with Mike’s wonderful simplification when $x>0$, we have
$$
I=-\frac{1}{x}-\int \frac{d x}{x^2 \sqrt{x^2+x+1}}
$$
Let $u=\frac 1x$ and then
$$
\begin{aligned}
\int \frac{1}{x^2 \sqrt{x^2+x+...
- 18.2k
1
vote
If $\alpha^3-\alpha+1=0$ then $\sqrt{3}\neq a\alpha^2+b\alpha+c$ for $a$, $b$, $c\in \mathbf{Q}$
Just to finalise what's been commented. Checking irreducibility of $f$ is maybe easier than you think!
$f$ is cubic, checking irreducibility amounts to checking no rational roots exist. Were $f$ ...
- 35k
1
vote
Accepted
Confused about the extraneous root of $\frac{x \sqrt{A^2 - x^2} + x}{x^2 - \sqrt{A^2 - x^2}}$
$\sqrt{A^2 - x^2} = -1$ has no real solution.
LHS is defined only when $A^2 - x^2 \ge 0$ in which case, LHS is non-negative and so, the equation has no real solution.
Thus, $x=0$ is the only real ...
- 388
1
vote
Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?
You got it wrong. She inspired her teacher to find this:
Suppose we want to find out that 21904 = 148²
We guess 99² and get 9801
We miss by |21904 - 9801| = 12103
Dividing by twice the guess we get ...
- 11
1
vote
Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?
It looks like a geometrical insight: the total surface equals the brown, plus two times one yellow plus the green: (the yellow ones are $1$ large).
The fact that a child (11 years old) has seen this, ...
- 1,623
1
vote
Solving $y+\sqrt{y^2-1}=e^x$ respect to $y$
I want to present a solution, albeit not as elegant as others; hopefully, you will feel you could have discovered it.
You have already tried using some trigonometric function to substitute $y$. I ...
- 3,372
1
vote
$f(x) = \sqrt x^{{\sqrt[3]{x}}^{\sqrt[4]{x},\cdots}}$ asymptotic?
With Mathematica there is a loglog plot
$$\text{ListPlot}\left[\text{Table}\left[\left\{x,(\log (\text{$\#$1}+1\&)\left((\log (\text{$\#$1}+1)\&)\left(\text{Fold}\left[\text{$\#$1}^{x^{\frac{1}...
- 1,131
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