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| stats | profile views | 586 |
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11h |
comment |
Way to have inutition about the shape of a curve? Try to think about long-term behavior, roots, and at least one point with nonzero output: often the $y$-intercept. As en example: $y=(x-1)^2e^{-x}$. In the long term, the exponential outweighs the quadratic. So as $x\to\infty$, this curve's $y$-values approach $0$. And as $x\to-\infty$, this curve's $y$-values approach $\infty$. The only roots are at $x=1$, and locally the function is $\approx c(x-1)^2$ there, so it's graph resembles a parabola near $x=1$ touching the $x$-axis. The $y$-intercept (and in fact every output) is positive. This is all enough information to sketch the curve roughly. |
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1d |
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Solve for an Ellipse Tangent to 2 Lines @flutefreak7 Also, you might be interested in GeoGebra where you can make slider bars for some of your variables, have circles and ellipses that depend on those things and then manipulate the slider bars. Just for starters, you could make a slider bar called "r" and enter "x^2+y^2=r^2" and then manipulate the bar to grow or shrink the circle. |
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1d |
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Solve for an Ellipse Tangent to 2 Lines Great! So, just to be clear, you could compute everything with $r=1$: you'd find values for all of the coordinates and for $a$ and $b$. Then if you wanted to make the circle larger by a factor of $r$, you could just make everything larger by a factor of $r$. This is a common trick to simplify calculations, and it is applicable in this case. |
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2d |
revised |
Solve for an Ellipse Tangent to 2 Lines added 407 characters in body |
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2d |
answered | Solve for an Ellipse Tangent to 2 Lines |
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2d |
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Puzzle identification and solving algorithm @noob I added a $2\times2$ example. Also, you should understand that order is irrelevant. Whether we push button A, then button B or vice versa, the net effect is the same. |
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2d |
revised |
Puzzle identification and solving algorithm added 1263 characters in body |
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2d |
answered | Puzzle identification and solving algorithm |
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2d |
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Puzzle identification and solving algorithm You may need to be much more explicit with how a numbered light is connected to the two numbered buttons that power it. Your first two lights use (25, 36) and (25, 55). If your third light used (36, 55) these three lights form an odd cycle that would prevent a solution. I know your third light is not using these buttons, but if there is an odd cycle somewhere else, there would be no solution. So I'm afraid that you need to give a fomrulaic pattern for how light $N$ relates to buttons $(f(N),g(N))$; either that or list everything out. |
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May 15 |
revised |
Conjecture regarding trapping rational numbers in some special intervals added 38 characters in body |
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May 15 |
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Conjecture regarding trapping rational numbers in some special intervals OK, now I think you are actually asking a question about a union of intervals. Sorry man - it's your question - I'll let you wordsmith it. |
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May 15 |
comment |
Conjecture regarding trapping rational numbers in some special intervals If I understand you, I would write this as: Let $b\in\mathbb{N}_{\geq3}$ and $\{x_j\}$ be a collection of $b-2$ rational numbers greater than $1$. Prove that there exists natural numbers $a$ and $c$ with $c\leq a$ such that for all $j$, that $\frac{ab-1}{(c-1)b+1}\geq x_j\geq\frac{ab-1}{cb-1}$. Note that this does not introduce $a$ or $c$ until a request is made to prove their existence. |
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May 15 |
comment |
Conjecture regarding trapping rational numbers in some special intervals You begin with "let $a\in\mathbb{N}$" and even use $a$ in the definition of $c$. Then you ask to prove the existence of $a$ subject to some condition. Do you see how this is a problem? Can you try to clarify the posing of the question? |
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May 15 |
revised |
Differentiation of $x^{\sqrt{x}}$, how? edited body |
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May 15 |
revised |
Differentiation of $x^{\sqrt{x}}$, how? added 466 characters in body |
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May 15 |
answered | Differentiation of $x^{\sqrt{x}}$, how? |
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May 15 |
revised |
Evaluation of a specific determinant. added 4 characters in body |
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May 15 |
answered | Evaluation of a specific determinant. |
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May 15 |
comment |
Approximation of alternating series $\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + …$ I always wonder why one would generally choose to estimate a series like this with a partial sum rather than averaging two consecutive partial sums. The error in $\frac{S_N+S_{N+1}}{2}$ is bounded by $\frac{1}{2}b_{N+1}$ and depending on the series, that could reduce the number of terms that are used by quite a bit. |
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May 15 |
comment |
Approximation of alternating series $\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + …$ You are trying to use integral test-based estimation on an alternating series. The integral test is only valid when $a_n=f(n)$ with $f$ a positive continuous decreasing function. $(-1)^n$ is not continuous, so you have no hope of integrating what you have written. Instead, you should look into alternating series test-based estimation, which is actually much simpler to execute. |
