Recent content by ialink

Hey Tiny Tim If i Would express δ(ε) I'd say δ(ε) = ε / |x+2|. Calculating δ with this function is possible for D = ℝ (for both ε and x) so a logical conclusion is that the limit exists that's what you mean right? but ε = 0.01 is given to find the appropriate δ. Something has to be assumed...

Using the product rule gives \frac{2}{3}x^{-1/3}*(x^{2}-4) + 2*x^{5/3} modifying this gives (try it): x^{-1/3}*(\frac{8}{3}x^{2} - \frac{8}{3}) For min/max equal both factors of this equation to 0 and off course check the domain. Does the expression exist at min/max?

Homework Statement Find limit L. Then find δ > 0 such that |f(x) - L| < 0.01 Limit when x approaches 2 of x^2 -3 Homework Equations 0 < |x-c| < δ 0 < |f(x) - L| < ε The Attempt at a Solution function is continuous at D = ℝ so limit is f(2) = 1 0 < |x-2| < δ and 0 < |x^2 -3...

Okay for instance if you expand (6+1)^2 you get 6^{2} + 2*6^{1} + 1 (equal to 6^{0}). The question was to prove that 7^{n} - 1 is devisable by six so if al components are a multiplification of six this is proven for the particular instance. Now if (6+1)^{2} = 6^{2} + 2*6^{1} + 1 then 7^{2} - 1...

Sigma I'm not familiar enough with. I think i know what to do. I'm going to do appendix E with is about sigma notation. I think i'll be much more able to solve this after having done this appendix. I've already done App A,C,D and they were all necesarry for chapter 1. I'm going to let this...

I've done some calculating: (6+1)^{1} = 6^{1} + 6^{0} \rightarrow 7^{1} - 1 = 6^{1} (6+1)^{2} = 6^{2} + 2*6^{1} + 6^{0} \rightarrow 7^{2} - 1 = 6^{2} + 2*6^{1} (6+1)^{3} = 6^{3} + 3*6^{2} +3*6^{1} + 6^{0} \rightarrow 7^{2} - 1 = 6^{3} + 3*6^{2} + 3*6^{1} (6+1)^{4} = 6^{4} + 4*6^{3} + 6*6^{2}...

No i haven't had that but figuring out it's pattern is not the only thing. In it's context you have to apply mathematical induction and that's the main goal of this an the coming 4 excercises. Is a reviewing chapter at the end of chapter one which has as main goal learing how to solve problems...

I mean that i first want to figure out the pattern. Induction comes after that

a^2 - b^2 = (a+b)*(a-b) but that's af far a i come. Writing it in anything like (a+b)^n*(a-b)^n returns large qeues of powers but i couldn't figure it out to get a^3 - b^3 for a starter. Can you give me some more information? grtz Ivar

Homework Statement Prove that 7^n - 1 is divisible by 6 Homework Equations The Attempt at a Solution I managed a solution were i use only the addition of 6^x were x is any postive integer. The solution i have workes until n = 3 but not after that. 7^n - 1 = 6^n + n*6^(n-1) +...

that's oke

Hurkyl already helped me check my answer which turned out to be right before you started posting so what makes you think i need an answer? I already proofed that h = \frac{p^2}{2p+24}. Maybe first read what the question is before fireing away. It's not that i don't appreciate help but i'd more...

Drawing the graph is always a good idea. You know the graph of y=x and i assume also the graph of 2*x^1/2. After that you can easily sketch x-2*x^2 and then you'll probably have a better idea were to look

With too simple i meant the same as you. It doesn't cover the question. I think we can agree on that grtz Ivar

I agree that it may be the solution for this particular case but not for all cases. The formula must cover all possibilities with one angle of 90 and the other two having a sum of 90. The simpeler the solution the better but this is too simple in my opnion.