Search results

3 ways. 1. Check with calculator \sin\frac{49\pi}{30}-\sin\frac{41\pi}{30}=0 and \sin\frac{53\pi}{30}-\sin\frac{37\pi}{30}=0 2. Use sum to product...

There have to be only 5 distinct answers. sin(49pi/30) and sin(53pi/30) are included because \sin\frac{49\pi}{30}=\sin\frac{41\pi}{30} and \sin\frac{53}{30}=\sin\frac{37\pi}{30} So it matters not if you list them as...

1 One might think it is undefined and if you try to do it in wolframalpha it can't do it. But if you want properties of matrices to be conserved it will be 1. For example if P=(A,0;0,Q), then |P|=|A||Q| and if A is empty then |P|=|Q| and so |A|=1.

I made a picture if it helps.

Just let \textstyle\frac{k}{n}=x and \textstyle\frac{1}{n}=\delta x As k goes from 1 to n, x goes from \textstyle\frac{1}{n} to \textstyle\frac{n}{n} which as n\rightarrow\infty means x goes from 0 to 1. And of course in the limit \delta x becomes dx

Do it as an integral...

Schaum‘s outlines are pretty good. I have hardcopy’s of them and there are so many that they will fill a whole shelf. But they are also available as ebooks. I also have the ebooks and maybe I’m old-fashioned but when I consult the Schaum outlines I almost always look at the hardcopies, not the...

Darth Vader saw the latest episode of Mandalorian replete with Christopher Lloyd and Jack Black and said Noooooo!:

Oh my goodness what is Star Wars coming too? In the remake of Return of the Jedi we saw the infamous “Nooooooo” Now we have Christopher Lloyd and Jack Black in Mandalorian S03E06? That was bad! I never thought Mandalorian could become a bad comedy act. But it almost got there with this...

Non-investigative Year 11 additions are at https://www.angelfire.com/ab7/fourunit/Cambridge-y11-e1-sol-nia.pdf Together with http://www.angelfire.com/ab7/fourunit/Cambridge-y11-e1-sol-cor.pdf and http://www.angelfire.com/ab7/fourunit/Cambridge-y12-e1-sol-cor.pdf this now reduced the list of the...

Here is a challenge: Redo EVERY divisibility proof you have ever done by induction, but without induction.

As with many divisibility proofs however it is probably better to do it without induction. 35^n+3\times7^n+2\times5^n+6=(7^n+2)(5^n+3) \begin{aligned}5^n+3&=5^n-1+4\\&=(5-1)(5^{n-1}+5^{n-2}+\cdots+1)+4\\&=4(5^{n-1}+5^{n-2}+\cdots+1+1)\text{ is divisible by 4 and}\end{aligned}...

1+3+2+6=12\text{ is divisible by }12\therefore\text{ it is true for }n=0. \text{If it is true for }n=k\text{ then }35^k+3\times7^k+2\times5^k+6=12M\text{ for an integer }M...

It is also true for n=6. Induction proof. \frac{6^6}{6!}=64.8>64=2^6\therefore \text{true for }n=6 \text{If true for }n=k, \frac{k^k}{k!}>2^k \text{and noting that }\frac{(k+1)^{k+1}k!}{k^k(k+1)!}=(1+\frac{1}{k})^k=1+k\cdot\frac{1}{k}+\sum_{r=2}^k{k\choose r}\cdot\frac{1}{k^r}>2 \text{then...

\lim\limits_{n\rightarrow\infty}(-1)^{n+1}\int_0^1\frac{t^n}{1+t}dt=\int_0^10dt=0 \text{This is because as }n\rightarrow\infty, t^n\rightarrow0\text{ for }0<t<1

It's kind of the same. a+ar+ar^2+ar^3+\cdots+ar^{n-1}=\frac{a(1-r^n)}{1-r} a=1,r=-t 1-t+t^2-t^3+\cdots+(-1)^{n-1}t^{n-1}=\frac{1-(-t)^n}{1--t}=\frac{1}{1+t}+(-1)^{n+1}\frac{t^n}{1+t}

\text{Suppose }x=-t \text{Noting that }1-x^n=(1-x)(1+x+x^2+x^3+\cdots+x^{n-1})\text { then} (1+t)(1-t+t^2-t^3+\cdots+(-1)^{n-1}t^{n-1})=1-(-t)^n \therefore1-t+t^2-t^3+\cdots+(-1)^{n-1}t^{n-1}=\frac{1-(-t)^n}{1+t}=\frac{1}{1+t}+(-1)^{n+1}\frac{t^n}{1+t}

You can apply integration to the second part thus avoiding Stirling's formula but for now I'll keep the Stolz–Cesàro theorem for the first part. Maybe find another way for that later. \begin{aligned}\text{By the Stolz-Ces\`aro theorem...

So how to do it without wolframalpha? \begin{aligned}\text{By the Stolz-Ces\`aro theorem...

e^{-1} which I got from wolframalpha.