Recent content by rakalakalili

So if hg is in H, then that implies g is in H, which implies that g-1 is also in H. A similar argument shows that g and g-1 are in K. So for each element g of HK, hk will be equal to hgg^-1k, and that is where the extra count come in. Thank you!

You have forgotten the chain rule when you did the product rule. The derivative of cos(2x) will be -2sin(2x) [Note the 2 in front, from the chain rule]. Other than that, your answer is the same as the book, based on the identity that cos(2u)=cos^2(u)-sin^2(u) (Alternatively you could have used...

What you showed was that the sequence converges to 0, not the series. The sequence has to converge to 0 for the series to converge, but it does not guarantee that it does. A good example of this is the harmonic series \sum_{k=1}^{\infty} \frac{1){k} The limit as k goes to infinity of 1/k is...

Homework Statement If G is a group, and H and K are subgroups of G, prove: |HK|=\frac{|H||K|}{|H\cap K|}Homework Equations The Attempt at a Solution A basic counting argument can be made by saying that HK consists of an element of H with an element of K, so you have |H||K| elements. What I...

Assuming you made a typo and your problem is the limit as t -> 3, then yes, what you did is correct. You can only ever cancel an expression with a variable (such as (t-3)) if you are sure that it is not going to be zero. In this case, we are taking the limit as t goes to 3, so (t-3) is never...

Homework Statement Hello, this is a problem from the practice test for the GRE subject test. For what value of b is the line y=10x tangent to the curve y=e^{bx} at some point in the xy-plane? A) \frac{10}{e} B)10 C)10e D)e^{10} E)eHomework Equations The Attempt at a Solution For the line to be...

That's perfect thank you! I had known that there was an infinite number of terms less than s+ \epsilon, but I did not connect that with making an interval, and simply choosing a term from each interval to form the subsequence. Thanks again.

One must always be careful when cancelling variables (as soon as you cancel, you assume your variable cannot be equal to 0). When solving for a in a^2 = 3a, you can look at it as the quadratic equation a^2-3a = 0. Then solve this using your favorite technique to solve for the roots of a...

So are the limits the same then, once you plug 3 in for a? Also, are you sure 3 is the only a that works? Are there any others?

Homework Statement Let (an) be a boundedd sequence, and define the set S= {x\in R : x < a_n for infinitely many terms a_n\} Show that there exists a subsequence (a_n_k)converging to s = sup S Homework Equations This is supposed to be a direct proof of BW using the LUB property, so no...

Look at (n-k+1)(n-k)!, can you simplify this product any?

Because you said that you never learned factorials let me further explain Bohrok's reply. 5! = 5*4*3*2*1, 6! = 6*4*3*2*1 and so on. Think of n! as multiplying all of integers from 1 up to n together. i! = i*(i-1)*(i-2)...*3*2*1. The product of all the integers from 1 up to i. (i+1)! =...

Have you tried writing the function as it's series representation about 0? Then the calculation you are looking for would be the coefficient in front of the z^100 term.