Recent content by QuantumP7

Aha! Thank you SO much! Awesome question and solution!

I'm very curious about this. I've been pouring over this solution because I think it's REALLY neat. So far, I've been able to follow: $$ \left( 1+ \frac{x}{n} \right)^{n} = e^{\ln \left( 1+ \frac{x}{n} \right)^{n}} = e^{n \ln \left( 1+ \frac{x}{n} \right)}$$ The series expansion for ##\ln...

Thank you, sir! Even though I misread part b, I knew that there was some deeper theory and a more elegant solution that I was unaware of. If no one else attempts part b within a few days, then I will redo it, myself. I look like a complete idiot in this thread, but I really am learning a lot.

5.(b) Show that there is an integer ##a \in \mathbb{Z}## such that 64959| ## (a^{2} - 7)## Answer: We can see that 3 divides 64959 because its digits add up to a multiple of 3 (6+4+9+5+9 = 33 = 3+3=6). (If necessary, I can explain why. Just let me know.) So, we need to find an integer a such...

##3^{2405} = 3^{5+2400} = (3^{5})(3^{2400})## ##3^{400} \equiv 1(mod 1000)## so ##3^{2400} \equiv 1(mod 1000)## Then, ##3^{2405} = (3^{5})(3^{2400}) \equiv (243*1)(mod 1000)##, so that the last 3 digits of ##3^{2405}## are 243.

5. a) Find the last 3 digits of ##3^{2405}## Answer: ##3^{2405} = \left( 3^{5} \right) ^{481} = \left( 3^{5} \right) ^{400+80+1}## The last 3 digits of ##3^{2405}## will then be ## \left(3^{5} \right)^{1} = 243##.

Aw. You're right. 4. Let x be a real number. Find ## \lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{x}{n} \right) ^{n} - e^{x} \right)## Answer: ## \lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{x}{n} \right) ^{n} - e^{x} \right)## ##= \left( \lim_{n \rightarrow \infty} n \right)...

4. Let x be a real number. Find ## \lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{x}{n} \right) ^{n} - e^{x} \right)## Answer: ##\lim_{n \rightarrow \infty} ln \left( n \left( \left(1 + \frac{x}{n} \right) ^{n} - e^{x} \right) \right)## ## = \lim_{n \rightarrow \infty} \left( ln(n)...

4. Let ##x## be a real number. Find ##\lim_{n \rightarrow \infty}n((1+\frac {x}{n})^{n} - e^{x})##. Answer: ##\lim_{n \rightarrow \infty}n((1+\frac {x}{n})^{n} - e^{x}) = \left( \lim_{n \rightarrow \infty}n * \lim_{n \rightarrow \infty}(1+\frac {x}{n})^{n} \right) - \left( \lim_{n...

Thank you SO much! I hadn't even thought to use ##\sqrt{n^{2} + 2n} = \sqrt{n^{2}} \sqrt{1 + \frac{2}{n}}##! Using my fraction, I somehow got that the limit of ##b_{n}## was -2. I see where my mistake was. When I used \frac{\frac{-2n}{n}}{\frac{n + \sqrt{n^{2} + 2n}}{n}} = \frac{-2}{1 +...

YES! That is the hint I needed! Thank you SO much!

Homework Statement Consider the sequence given by b_{n} = n - \sqrt{n^{2} + 2n}. Taking (1/n) \rightarrow 0 as given, and using both the Algebraic Limit Theorem and the result in Exercise 2.3.1 (That if (x_n) \rightarrow 0 show that (\sqrt{x_n}) \rightarrow 0), show \lim b_{n} exists and find...

I asked a Putnam Fellow this question. He said that the best way to do really well on the Putnam is to practice. Go over the old questions, and practice a lot! I'm going to do this all of 2014, and take the Putnam in December 2014. I'll let everyone know how it turns out.

I'm curious. What WOULD impress Harvard or MIT or the other top math programs?

I became fully deaf about a year and a half ago. I've always had problems with my hearing and severe depression, so no degree yet. I've been studying finance so that I can try to make some money and get some cochlear implants (Medicaid in my state doesn't pay for it), and get off of SSI. I...