14
votes
Accepted
How can I learn how to solve hard problems like this Example?
Technique : rationalization
When we are given $A/B$ or $1/B$ , we try to eliminate the Denominator.
That will involve "conjugates" , that is , When Denominator is $x+\sqrt{y}$ , we could ...
- 12.3k
5
votes
How can I learn how to solve hard problems like this Example?
And guess what, these problems are hard for me.
Should I spend more time on hard ones before giving up?
Should I memorise solutions by heart to get techniques?
What could I do if I get stuck?
I am ...
- 17.3k
4
votes
How can I learn how to solve hard problems like this Example?
Let's start with some tough love:
If you gave up after half an hour, you didn't really try.
Prem and others have already posted some concrete tips on how this particular problem could be approached, ...
- 9,892
4
votes
Accepted
Solutions to $(f(x)-f(y))^3=f\left(x^3\right)-f\left(y^3\right)$
I leave my proof, let me know if there are some incorrect reasonings please.
The globally constant functions satisfy our equation, so we can suppose there are $x$ and $y$ such that $f(x)\neq f(y)$. We ...
- 8,911
4
votes
Integration of a very hard expression
For all $n \in \mathbb {N}$, see
$$\begin{align} \int_{0}^{\infty} {t}^{n} {e}^{- t} \text {d} t & = - \int_{0}^{\infty} {t}^{n} \cdot \frac {\text {d}}{\text {d} t} {e}^{- t} \cdot \text {d} t \\ ...
- 1,472
3
votes
Accepted
How to prove $(x+y)|a|^2 |b|^2 - x |a|^2 - y |b|^2 = - (x|a|^4 + y |b|^4)$?
Use the constraint $|a|^2 + |b|^2 = 1$:
$$(x+y)\,|a|^2 |b|^2 - x |a|^2 - y |b|^2
\;=\; (x+y)|a|^2 |b|^2 - \Big(x |a|^2 +y |b|^2\Big)\Big(|a|^2 + |b|^2\Big)$$
then multiply out & simplify.
- 6,528
2
votes
How to make the largest square possible from smaller squares
Here's a 7X7 square
8X8 square can't be done
- 147
2
votes
Accepted
Is $(\mathbb{Q},0,+,<)$ isomorphic to $(\mathbb{Q}^{+},1,\cdot,<)$? ("Logic in Mathematics and Set Theory" by Kazuyuki Tanaka and Toshio Suzuki)
A positive rational number can be written uniquely
$$q=\prod_{p\in\mathcal P}p^{r(p)}$$
where $\mathcal P$ is the set of prime numbers and $r:\mathcal P\to\Bbb Z$ is of finite support (i.e. $r(p)=0$ ...
- 39.7k
2
votes
Accepted
Blitzstein, Problem number 7, Stat 110 Homework 4 Solutions, Fall 2011
You can define a random variable $Y_j:= \max_{1\leq i \leq j} X_i$. Then $\{ I_j=1 \}=\{ Y_j =X_j \}$.
Notice also that $ \cup_{\ell=1}^j \{ Y_j =X_\ell \} $ is the whole event space. Since $X_1,...,...
- 7,805
1
vote
Accepted
Problem E1 in Engels's "Problem Solving Strategies"
The difference between $x_{n + 1}$ and $y_{n + 1}$ gets smaller as $n$ grows because it's less than half the difference between the previous $x$ and $y$ coordinates.
This is the key. In general, just ...
- 8,679
1
vote
Is $(\mathbb{Q},0,+,<)$ isomorphic to $(\mathbb{Q}^{+},1,\cdot,<)$? ("Logic in Mathematics and Set Theory" by Kazuyuki Tanaka and Toshio Suzuki)
You can try to prove that $\mathbb{Q}^{×}$ is isomorphic to the direct sum of a cyclic group of order 2 and a free abelian group with countably many
generators, and $\mathbb{Q}^{+}$ is not a direct ...
- 119
1
vote
Is $(\mathbb{Q},0,+,<)$ isomorphic to $(\mathbb{Q}^{+},1,\cdot,<)$? ("Logic in Mathematics and Set Theory" by Kazuyuki Tanaka and Toshio Suzuki)
Let $F(1)=q=\frac{a}{b}\in\mathbb{Q}^{+}$ and $a,b\in\{1,2,3,\dots\}$ and $\gcd(a,b)=1$.
Let $n\in\{1,2,3,\dots\}$.
Then, $F(\frac{1}{n})=\sqrt[n]{\frac{a}{b}}=\frac{l}{m}$ for some $l,m\in\{1,2,3,\...
- 1,138
1
vote
Which of the equations has solutions in $\mathbb{Z}$?
HINT.-Because of by q.r.l. $(\dfrac{-1}{p})=(-1)^{\frac{p-1}{2}}$ which is negative for $p=3,7$, we have that $-1$ is not a square modulo $p$ for $p=3,7$. It follows
$3x^2+5y^2=7z^2$ and $3x^2+7y^2=...
- 30.3k
1
vote
Russian roulette with re-spin intuition for asymmetric solution
Each run is independent in a certain abstract sense.
We can imagine an indefinite sequence of spins of the cylinder,
always with one live bullet and pulling the trigger after each spin.
In this ...
- 100k
1
vote
How to prove $(x+y)|a|^2 |b|^2 - x |a|^2 - y |b|^2 = - (x|a|^4 + y |b|^4)$?
Hint: $$-x|a|^2-y|b|^2=(-x|a|^2-y|b|^2)(|a|^2+|b|^2)$$ because by assumption $|a|^2+|b|^2=1$.
- 41
1
vote
Then the number of solutions of $| x^ 2 − [ x ] | = 1$ is?
There is a solution, which does not imply any graphical ideas. First of all, your equality is equivalent to: $x^2-[x]=1$ or $x^2-[x]=-1$, from where we will write $[x]$ as $x-frac(x)$, where $frac(x)$ ...
- 65
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