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20h
comment How to prove $ \sum\limits_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$ using induction?
Its denominator is exactly what is needed to make the middle fraction have the same denominator as the fraction that follows. The numerator is just so that we are multiplying by $1$. It is pretty important to understand how factorials work for this. You need to see that these two fractions have basically the same denominator, with the middle one having an additional factor of the next integer, $(c+1)+1$.
20h
answered How to prove $ \sum\limits_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$ using induction?
1d
comment Can $R \times R$ be isomorphic to $R$ as rings?
The bijection from $R\times R\to R$ would be something like $(x_1,x_2,\ldots;y_1,y_2,\ldots)\mapsto(x_1,y_1,x_2,y_2,\ldots)$.
Apr
14
comment Why does the integral of absolute value function not return actual area?
GeoGebra. But graphing is not what led to me seeing the discontinuity. The denominator shows that the proposed antiderivative is discontinuous at $\pm1$, so I though about $1\pm\epsilon$ for $x$ and how the output would behave, and algebraically worked out the one-sided limits. I added that explanation between the pictures. I should have included it originally.
Apr
14
revised Why does the integral of absolute value function not return actual area?
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Apr
14
comment Why does the integral of absolute value function not return actual area?
@CShreve OK, pictures added.
Apr
14
revised Why does the integral of absolute value function not return actual area?
added 722 characters in body
Apr
14
comment Why does the integral of absolute value function not return actual area?
It's not continuous. I'll add pictures.
Apr
14
comment Determining the UMVUE for a discrete scaled uniform sample.
In hindsight, I think this was an easier computation than what you are trying because it treats the problem as sampling rather than with independent variables. The binomial coefficients and the identities used make things easier to sum than the powers of expressions that you have.
Apr
14
revised Determining the UMVUE for a discrete scaled uniform sample.
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Apr
14
answered Determining the UMVUE for a discrete scaled uniform sample.
Apr
14
revised Determining the UMVUE for a discrete scaled uniform sample.
edited body
Apr
14
comment Determining the UMVUE for a discrete scaled uniform sample.
Just a comment: in the real German tank situation, the $X_i$ are not independent, since if say $X_1=42$, then no other $X_i$ can be $42$. If $n$ is much smaller than $N$, you could pretend the $X_i$ are independent. Maybe none of this matters depending on the analysis.
Apr
13
answered Why does the integral of absolute value function not return actual area?
Apr
13
revised Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$
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Apr
13
answered Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$
Apr
13
comment Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$
Once you are down to $p(p+q)\equiv3\pmod{5}$, then it's clear that $p$ can be anything nonzero mod $5$, and there is a $q$ mod 5 that pairs with it. So an analysis mod 5 only is not going to get any farther.
Apr
11
answered What does this notation mean: $\tiny\left(\begin{matrix} y2-N \\ d_f -c, d_f-c \end{matrix}\right)$?
Apr
11
comment Why is the integral of $\int {1 \over {1+ \sqrt{x}}} dx$ not the same as for $\int{1 \over u}$?
Is $\int\frac{1}{x^2}\,dx$ equal to $\ln(x^2)+C$? You should untrain yourself from seeing a reciprocal function and jumping to $\ln$ for the antiderivative. That's only true for the reciprocal a linear function whose slope is $1$.
Apr
11
revised Why is the integral of $\int {1 \over {1+ \sqrt{x}}} dx$ not the same as for $\int{1 \over u}$?
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