Recent content by 3.14159potato26

What's with all the notation ambiguity? First off, if read conventionally, it would mean (tana) to the power of -1 [tan(n+1)a-tan(na)]. So, is it that or the multiplication of (tana) to the power of -1 with the other term [tan(n+1)a-tan(na)]? Another thing is the meaning of (tana)^-1. No...

Since the question doesn't explicitly say that you have to use induction to solve it (although induction is much simpler), you can also use an analytical method to prove this result. [Syntax: a | b means b is divisible by a] Now, from the question, 2 | n, and we need to prove that 8 | n^2 + 2n...

A different method that doesn't involve simultaneous equations is by observing that the value of the x-intercepts are already given, i.e. roots are x = 0,4. Therefore, Sum of roots = 0 + 4 = 4 Product of roots = 0 * 4 = 0. By definition, f(x) = x^2 - (Sum of roots)x + (Product of roots) f(x) =...

Well, I dunno about the previous solution (not that I have anything against it, really) but here's how I look at it: To determine the coefficients of terms in an nth degree polynomial, you need at least (n+1) distinct equations in terms of the coefficients. This rule is quite obvious. For...

Um...yeah, but consider the following: Let A be the (non-symmetric) matrix: {0 4} {1 0} The characteristic polynomial is (t - 2)(t + 2), therefore t = -2,2. Eigenvectors are given by the equation: {t -4}{a} = 0 {-1 t}{b} Solving it generally gives the eigenvectors as x = (-2b,b) and (2b,b)...

xD....i'm not up to that level yet, have mainly been concentrating on algorithms (i.e. direct/iterative methods for solving systems of linear equations), but thanks anyways, i'll try to look through it when i'm free.

A symmetric matrix is one that is equal to its transpose, i.e. A = A^T. Consider the following 3-by-3 matrix: { 0 0 1 } { 1 0 0 } { 0 1 0 } It can be seen that this is not a symmetric matrix. However, by interchanging rows (2) and (3), the matrix: { 0 0 1 } { 0 1 0 } { 1 0 0 } is obtained. This...

Its asking you to prove that 1/1! + 1/2! + 1/3! + 1/4! + .... < 103/60 + 1/((e^5)*(e-1)) using the given formula 1/n! > 1/e^n; n > 5. Easier to prove LHS < RHS rather than RHS > LHS.

Nope, there is no end. The whole point of the ... means it goes on to infinity. However, since it is a convergent infinite series, it has a finite value, which you don't need to calcuate, but just prove that LHS < RHS, using the given formula.

In that case, you could just add up the first few terms, i.e. the first 5 terms to get a value > 1.45, then claim because 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 < 1 + 1/2^2 + 1/3^2 + ... + 1/99^2, 1 + 1/2^2 + 1/3^2 + ... + 1/99^2 > 1.45.

f(x)= [4/(x-1)] - 2 ; x<1. Let f(f^-1(x)) = x x = [4/(f^-1(x)-1)] - 2 x + 2 = 4/(f^-1(x)-1) f^-1(x) - 1 = 4/(2+x) f^-1(x) = 1 + 4/(2+x) f^-1(x) = (6+x)/(2+x) New question: Given that 1/n! > 1/e^n; n > 5, prove that 1/1! + 1/2! + 1/3! + 1/4! + .... < 103/60 + 1/((e^5)*(e-1)).

Yes it is, draw the graph, take out the retangles and you'll find that its true. Its done by considering the areas under the graph (which is what the question asks you to do), and since one is obviously smaller than the other, so yeah, as far as i know, its valid.

Well, to prove that 1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2, consider the graph of 1/x^2. From the graph, I{1->100} 1/x^2 < 1 + 1/2^2 + 1/3^2 + ... + 1/99^2, that is taking the higher/left value of the function values as the retangle height, i.e. if function values are 1/n^2 and 1/(n+1)^2...

Um....for n = 1, you get 1.45 < 1, which is false. So does 1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/n^2 <1.99 say that it is true for any arbitary n, or are you talking about the limiting sum of 1 + 1/2^2 + 1/3^2 + ...+1/n^2?

When n = 1, LHS = 1 RHS = 2 - 1/1 = 1 Since LHS <= RHS, it is true for n = 1. Assume that it is true for n = k, i.e. 1 + 1/2^2 + 1/3^2 + ...+1/k^2 <= 2-1/k Consider 1 + 1/2^2 + 1/3^2 + ...+1/k^2 + 1/(k+1)^2 <= 2 - 1/k + 1/(k+1)^2 = 2 - 1/k + 1/(k+1)(k+1) <= 2 - 1/k + 1/k(k+1) ---(because k+1...