2h
comment Why is this wrong? Simple algebra
And, why are people downvoting this?
2h
comment Why is this wrong? Simple algebra
You're essentially saying ${1\over {1\over a}+{1\over b}}={a+b}$. Try that with $a=b=1$; you get $1/2$ on the LHS and $2$ on the RHS.
2h
comment Why is this wrong? Simple algebra
The reciprocal of the denominator is not as you suggest. Do the addition downstairs first, then multiply by the reciprocal.
4h
comment Prove this series is convergent.
And, for what it's worth, you can rearrange conditionally convergent series, as long as the rearangement is done in "blocks" with the size of the blocks bounded by some $M$.
4h
comment Prove this series is convergent.
It is correct. Nothing was rearranged. Just simplify $\sum_{i=1}^n a_i$, where $a_i$ is the $i$th term of your series.
4h
comment Prove this series is convergent.
Compute the partial sums; simplify by grouping terms as suggested above. You'll see why the answers below are correct.
4h
comment Prove this series is convergent.
Write it as $\color{maroon}{-{2\over 2^2}+{1\over 2^2}}\color{darkgreen}{-{3\over3^2}+{2\over3^2}}-{4\over4^2}+{3\over4^2}-\cdots$. Do you see how to handle it?
4h
comment Prove this series is convergent.
Oh, I misread it.
7h
comment Is there a countable basis for the finite complement topology on the natural numbers?
This is a duplicate, btw.
7h
comment Is there a countable basis for the finite complement topology on the natural numbers?
No. The closed sets are the finite subsets of $\Bbb N$. This collection is countable. So, the collection of open sets is countable as well.
7h
comment Is there a countable basis for the finite complement topology on the natural numbers?
What's the cardinality of the collection of closed sets?
7h
comment Let $k \geq 2 \in \mathbb{N}$ Show that each of $k! +2, k! +3, ... , k! +k$ is composite.
If $k=2$, you're only looking at $2!+2$. For $k=3$, you'll look at $3!+2$, $3!+3$. For $k=4$, you'll look at $4!+2, 4!+3, 4!+4$. ...
8h
revised Find maximum area of triangle defined by tangent line to $y=e^{-x}$
edited tags
9h
comment Proof that the graph of a linear function and its inverse cannot be perpendicular.
It looks fine; except you should state at the outset that you're assuming the two graphs are perpendicular.
9h
comment Is it true that they really chose dashed lines for > and < and solid lines for $\ge$ and $\le$ in linear inequalities for graphing?
Sounds reasonable.
10h
comment Finding linear transformations.
What are your choices for the images of the basis elements $(1,0,0)$, $(0,0,1)$?
10h
comment show that the function below is linear
See this.
10h
comment Both $F$ and $C$ are closed sets but their sum $F+C$ is not closed.
For $\Bbb R$, see this. Similar examples can be found in $\Bbb R^n$.
11h
comment Show that $\int^\infty_0$ $\int^\infty_0$ sin($x^2$+$y^2$) dxdy value is $\frac{\pi}{4}$
I don't think the polar coordinate approach is valid, since it assumes $\int\kern-3pt \int_C \sin(x^2+y^2)\, dA$, where $C$ is the first quadrant, exists. But it doesn't, as computing $\lim\limits_{n\rightarrow\infty} \int\kern-3pt\int_{D_n} \sin(x^2+y^2)\,dA$, where $D_n$ is the portion of the circle $x^2+y^2<n^2$ in the first quadrant, shows. (Then, one computes $\int_0^{\pi/2}\int_0^n r \sin r^2\,dr\,d\theta={\pi\over4}(1-\cos n)$.)
11h
comment Show that $\int^\infty_0$ $\int^\infty_0$ sin($x^2$+$y^2$) dxdy value is $\frac{\pi}{4}$
You get $\int_0^{\pi/2}\int_0^\infty r\sin r^2\,dr\,d\theta={\pi\over 2}\int_0^\infty{1\over2}\sin u\,du$. Perhaps I'm off...
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