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Re: Several Variable Calculus Correct! This is the Fourier series of a discontinuous function (a square wave) on [-pi, pi) say. What do you know about Fourier series of functions with jump discontinuities? (If you haven't seen it before, you may want to read about this...

Re: Several Variable Calculus Do you know where this series comes from? If so, you can use that to help prove it.

Re: MATH1131 help thread The difference between "the substitution method" and inspection is that if you do it by inspection, you don't write out the substitution, you just write down the answer.

Re: MATH1131 help thread $\noindent The derivative of $1 + x^{2}$ is sitting there, so an antiderivative is $\frac{1}{4}\left(1 + x^{2}\right)^{4}$.$ $\noindent Use a mental substitution of $u = 1 + x^{2}$.$

Try and understand the problem and translate it into mathematics.

$\noindent The answers are both valid antiderivatives because they differ by a constant. The first answer becomes$ $$\begin{align*}\frac{1}{4}\ln \left(4(x+1)\right) &= \frac{1}{4}\left(\ln 4 + \ln (x+1)\right) \quad (\text{log law:} \ln (AB) = \ln A + \ln B) \\ &= \frac{1}{4}\ln 4 +...

Why not post your working, and maybe someone can spot your mistake. In writing out your working here, it's possible that you may spot a mistake too.

Re: Several Variable Calculus In other words, take this graph ( http://www.wolframalpha.com/input/?i=graph+(1-6%7Cx%7C%5E2%2B4%7Cx%7C%5E3)+from+-1+to+1 ) and imagine repeating that same picture on [1, 3], and [3, 5] and [5, 7] and ..., and also on the negative side (so on [-3, -1], [-5, -3]...

Re: Several Variable Calculus The function has been defined on [0, 1] by that formula, and then we are told f(-x) = f(x) for all x and f(x) = f(x+2) for all x, so this means the function is an even periodic function with period 2, and on [0, 1] it looks like 1 - 6x^2 + 4x^3. In other words...

Expand the numerator using the formula for (A + B)^2, and then simplify by splitting the fraction. This will leave us with terms that can be integrated using the power rule.

Re: Several Variable Calculus $\noindent One quick test is to just check the constant term first (since your answer and the given answer have different constant terms; your one is $-1$ and the given answer's is 0). $\noindent The constant term in a Fourier series is the \emph{average value}...

Are you sure those should be cos(x) and sin(x) instead of something like cos(theta) and sin(theta)? They wouldn't be straight lines as you've written them.

Re: Several Variable Calculus $\noindent One way you can do it is to show via calculus that $n^{\alpha}x^{n} (1-x)$ has maximum value when $x = \frac{n}{n+1}$ for $x\in [0,1]$, and thus find this max. value, and show it goes to $0$ as $n\to \infty$.$

Re: Several Variable Calculus $\noindent Let $\alpha < 1$ be fixed, and $f_{n}:[0,1]\to \mathbb{R}$ is given by $f_{n}(x) = n^{\alpha}x^{n}(1-x)$. That this converges to the zero function on $[0,1]$ uniformly means that for each $\varepsilon > 0$, there exists $N$ such that \textbf{for all} $x...

Re: Several Variable Calculus What was your progress so far?

What was your progress on this problem?

No. The inverse should involve a square root.

Yes

Try and solve that equation for x in terms of y. Also note that the domain of the inverse will be the range of the given function.