meta user
network profile
| bio | website | |
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| age | ||
| visits | member for | 1 year, 3 months |
| seen | 11 hours ago | |
| stats | profile views | 1,160 |
"You have enemies? Good. That means you've stood up for something, sometime in your life" - Winston Churchill
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May 6 |
comment |
If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. My +1 and Thanks. |
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May 6 |
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If $b^2$ is the largest square divisor of $n$ and $a^2 \mid n$, then $a \mid b$. @BillDubuque Recommend the long line to be split as $$ \begin{align*} a^2 | b^2c \Rightarrow a^2 | (bc)^2 &\Rightarrow a^2 | a^2c, abc, b^2c \\ &\Rightarrow a^2 | (a,b)^2c \Rightarrow (a/(a,b))^2 | c \end{align*} $$ My +1 for this answer. |
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May 6 |
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Result of the product $0.9 \times 0.99 \times 0.999 \times …$ @PaulManta this converges to $0.89$, but you have to think how to show that. |
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May 6 |
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Result of the product $0.9 \times 0.99 \times 0.999 \times …$ @PaulManta I changed the \cdots to \times because it was confusing with the numbers themselves with decimal values. |
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May 6 |
revised |
Result of the product $0.9 \times 0.99 \times 0.999 \times …$ edited title |
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May 6 |
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Result of the product $0.9 \times 0.99 \times 0.999 \times …$ How about $(1-\frac{1}{10})(1-\frac{1}{100})...= \prod_{i=0}^{\infty}(1-\frac{1}{10^i}) $ |
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May 6 |
awarded | Necromancer |
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May 5 |
awarded | Nice Answer |
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May 5 |
comment |
Find the sum of $\sum 1/(k^2 - a^2)$ when $0<a<1$ @N3buchadnezzar your expected answer has an error. Checked against Wolfram Alpha's snipurl.com/23dq2hp and instead of plus sign, it shows minus sign. Follow J.M's hint, your might approach the answer. |
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May 5 |
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If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. @dato, the original question was prove $1-xy$ is a perfect square, so there is a proof required, not solutions to $x,y$. |
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May 5 |
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If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. I think I do understand both your comments. Bill likes to welcome new users, and Patrick was right to mention that people who seek help should not just post a question without proper introduction. I think MSE should stop unregistered users to just post a question and vanish in thin air. |
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May 5 |
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If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. @PatrickDaSilva Thanks. |
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May 5 |
answered | If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. |
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May 5 |
comment |
If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. Did you at least find trivial ones, like $x=y=0$ then $1-xy=1=1^2$ ? |
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May 5 |
answered | Funny identities |
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May 5 |
answered | Integration: area enclosed by graph of $x^4 + y^4 = 1$ |
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May 4 |
answered | Why is fibonacci coding useful? |
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May 4 |
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Statistics : Where did this degree of freedom formula for the T distribution come from? Maybe stattrek.com/estimation/difference-in-means.aspx?tutorial=stat can help. Under the subsection titled "If you use a t score, you will need to compute degrees of freedom (DF)." |
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May 1 |
answered | Absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ |
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May 1 |
comment |
Integration of $\int_0^{\pi/2} \frac{1}{\tan^2(x)}dx $? The integral does not converge $-\cot x -x$ from $0 $ to $\frac{\pi}{2}$ |
