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Lol I found the mistake. It was a mistake on the paper when I wrote it down, but I posted the right equation.....

imatfaal we can use the equation I wrote down to figure out x1 or x2 is equal to zero when we have the maximum. From basic understanding we know the two critical points are going to be +/-5 and +/-3, but they will both be positive because the max is squared for each variable. I just don't understand how to finish this off with lagrange. I logically can figure out the answer.

Lagrange Problem max [math] y = x_1^2 + x_2^2[/math] such that [math] \frac{x_1^2}{25} + \frac{x_2^2}{9} -1 =0 [/math] [math] \frac{df}{dx_1} = 2x_1 + \frac{2x_1\lambda}{25}[/math] (1) [math] \frac{df}{dx_2} = 2x_2 + \frac{2x_2\lambda}{9}[/math] (2) [math] \frac{df}{d\lambda} = \frac{x_1^2}{25} + \frac{x_2^2}{9} - 1 [/math] (3) Dividing equation [math] \frac{1}{2} [/math] [math] \frac{x_1}{x_2} = \frac{9*x_1}{25*x_2} [/math] As you can see the x_1 and x_2 just cancels out. Do you know what the issue is? I tried using a quadratic, but it still has the lambda in it, so I cannot solve it that way. How can I solve this one using the Lagrange?

[math] y = cx^4 \Rightarrow ln{y} = ln{cx^4} \rightarrow ln{y} = ln{c} + 4ln{x} \rightarrow ln{y} = 4ln{x} + c [/math] I don't understand how the four gets inside the ln to become [math] ln{4y} [/math] Thank you for the help. Cheers.

Solve the following differential equation by separation of variables: [math] \frac{xdy}{dx} = 4y [/math] My solution is: [math]\int \frac{dy}{4y} = \int \frac{dx}{x} \Rightarrow ln{\mid 4y \mid} = ln{\mid x \mid} + c [/math] The book's solution is: [math] y = cx^4 [/math] Can anyone show me how they came up this this solution? Thank you.

X is a n*k matrix, k < n X is of full rank k (full column rank) X'X is of full rank and therefore invertible [math] P_x = X(X'X)^{-1}X'[/math] Show that [math]P_x[/math] is symmetric and idempotent. I figured out how to show it is idempotent. Here is my attempt to show it is symmetric: [math] (P_x)' = (X(X'X)^{-1}X')' = X'[(X'X)^{-1}]'X = X'(XX')^{-1}X[/math] Maybe someone has a sheet for all the algebra rules for linear algebra that would be helpful?

I am trying to solve [math] \frac{dx}{dt} = Rx(t)(1 - \frac{x(t)}{K} [/math] knowing [math] x(0) = K/N [/math] I got to [math] dx = [Rx(t) - R(x(t))^2]dt [/math] and then I am unsure about how I should do the integration of the second term of the right hand side. The solution is [math] x(t) = \frac{K}{1 + (N - 1)e^(-Rt)}[/math]

The third part was wrong. Here is the correct version: [math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}x'(b)^2 - \frac{1}{2}x'(a)^2 [/math]

[math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}\frac{dx}{dt} - \frac{1}{2}\frac{dx}{dt} [/math] What I am having trouble is understanding is going from the second step to the third step. I don't understand how you can integrate with respect to dx and still integrate [math] \frac{d^2x}{dt^2} [/math]

The questions asks to use mathematical induction to prove the truth of the follow for [math] n \geq 1 [/math] [math] a^n - b^n [/math] is divisible by [math] a - b [/math] for any integers a,b with [math] a-b \neq 0 [/math] Proof Attempt: [math] n = 1[/math] [math] a^n - b^n = a - b [/math] [math]\frac{a - b}{a - b} = 1 [/math] Now suppose that [math] n = k [/math] and the statement is true for [math] n = k [/math] We must show that the statement is true for [math]n = k + 1[/math] I tried using long division as it was suggested as a hint to consider the remainder: [math] a^{k+1} - b^{k+1}[/math] divided by [math] a - b [/math] gives a remainder of [math] -b^{k+1} + ba^k [/math] Any suggestions? Is this a historical proof by chance so I can see how it was done? Merged post follows: Update of attemptUsing the algebraic expansion: [math] a^n - b^n = (a - b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1} [/math] but for n = k + 1 solved the problem.

I am doing a Mathematical Induction out of the book with solution. I don't see how the LHS is equal to the RHS. [math] 1^2+2^2+3^2+...+(k+1)^2 = (1^2+2^2+3^2+...+k^2)+(k+1)^2 [/math] Expanding [math] (k+1)^2 = k^2 + 2k +1 [/math] What am I missing?

Of course. Thank you both for reminding me.

Why is [math] \lim_{t\to0} \frac{\sin(t) }{t} = 1 [/math]

I am trying to integrate [math] \int\frac{dx}{x\sqrt{x^6 -25}} [/math] I tried using algebra to change it so it easier to use substitution to complete the integration. Here what I did: [math] \int\frac{dx}{5x\sqrt{(\frac{x^3}{5})^2 -1}} [/math] How can I approach this problem? This is just a problem out of my book for practice. I do have the solution to the problem.

I have no idea how to find the real part of [math] z = e^{(1-i\frac{pi}{3})} [/math]

This is a question I have worked on in Stats and my answer does not line up with the solution manuals. [math] P(A' \cap B' \cap C) = P© - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) [/math] My answer there is a negative before [math] P(A \cap B \cap C) [/math]

I am trying to prove the follow two questions without using the differentiation formulas for inverse trig. or inverse hyperbolic functions. [math] \arcsin(u) = \arctan(\frac{u}{\sqrt{1 - u^2}}) [/math] I have been looking at it as [math] \frac{opp}{adj} = \frac{opp}{hyp} [/math] but something in my steps is off. I am also trying to solve [math] \frac{d}{dx}[\tan(\arcsin(x))] = \frac{1}{(1 - x^2)^\frac{3}{2}} [/math]

I am just doing a simple integral using substitution and when I check it by taking the derivative it not coming out as the same answer. I am taking [math] \int \frac{x}{(2+x^2)^2} [/math] and getting [math] \frac{-1}{2(2+x^2)} [/math]. When I take the derivative of that function I get [math] \frac{4x}{(4+2x^2)^2} [/math]. Where is my mistake?

I am working on Taylor Series and I believe one of the derivatives in the solution manual is wrong. I am taking the derivative of [math] f(x) = \frac{x}{(1-x^2)^{(3/2)}} [/math] I believe it is: [math] f'(x) = \frac{1+2x^2}{(1-x^2)^2} [/math] but the answer key says it is: [math] f'(x) = \frac{1+2x^2}{(1-x^2)^{5/2}} [/math] Can you check my math please.

How do I go about finding all the solutions to [math]x^4-8x^2+17=0[/math]?

Thank you for the help, as I am sure you noticed I used L'Hopital's rule to solve it. Though how do I do it with binomial expansion?

This is where I am at: Using/assuming the Continuity of f(x) [math] a_n = \left( 1 +\frac{1}{n} \right)^{n} = e^{x\ln(1 + \frac{1}{x})} [/math] [math] \lim_{x\to\infty} 1 + \frac{1}{x} = 1 [/math] therefore [math] e^{x\ln(1)} = 0 [/math] So I am making a mistake somewhere but I am not sure where. I should mention this is just a practice problem and not a homework assignment, but this seemed the best place to post the question. I understand it one of the standard definitions, but I want to prove it for my own knowledge. I am not using anything else as a definition of e. How could I do a binomial expansion as it is an infinite sequence? Here how to do it with L'Hopital's rule: [math] e^{x\ln(1 + \frac{1}{x})} [/math] [math] \lim_{x\to\infty} [/math][math] \ln(\frac{1 + \frac{1}{x}}{\frac{1}{x}} ) = 1[/math] therefore [math] =e^1 = e[/math] Anyone know how to do it another way?

Sorry about the confusion I believed that [math] a_n [/math] was one of the standard notions to mean it is an infinite sequence. Assuming that fact how do I prove it?

How do I show that [math] a_n = \left( 1 +\frac{1}{n} \right)^{n} = e [/math]

I am working in infinite sequences and I forgot something simple. What the rules for taking the derivative for a function such as this: [math] \left(\frac{3}{2} \right)^{x} [/math]