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This problem comes from one of my problem books, but the solution is not provided. The final answer is stated to be $\ln^2 2$, but I am struggling to understand how to evaluate the limit: $$\lim_{n \to \infty} \sum_{k=n+1}^{2n} \left( 2\sqrt[2k]{2k} - \sqrt[k]{k} - 1 \right)$$ I tried to rewrite the term inside the summation as: $2\sqrt[2k]{2k} - \sqrt[k]{k} - 1 = 2(\sqrt[2k]{2k} - 1) - (\sqrt[k]{k} - 1).$ I attempted using $x^x = 1 + x \log x + \frac{1}{2} x^2 \log^2 x + O(x^3 \log^3 x)$ However, this led to a rather complicated summation, and I got stuck trying to simplify it further.

I made a Geogebra plot about the problem. The problem from a 9th grader exam. https://www.geogebra.org/calculator/jdbrkgu6 As long as the 3 conditions hold, it seems that the $\angle \alpha$ is always a right angle. But I could not prove it, can anyone solve it, thanks.

In my analysis II class I was given the following question: Let f : (a, b)→ R for some a < b. f is differentiable at c ∈ (a, b) if and only if there exists some function L on (a, b) continuous at c such that for all x ∈ (a, b), f(x) − f(c) = L(x)(x − c). Use this result to prove that if f is defined on (a, b) for some a < b and f is differentiable at c ∈ (a, b), then f is continuous at c. This is what I have so far, if posisble can I get some feedback on my proof, thank you: Assume f is defined on (a, b) for some a < b and f is differentiable at c ∈ (a, b). Since f is differentiable there, there exists some function L on (a,b) continuous at c s.t. for all x ∈ (a, b), f(x) - f(x)= L(x)(x-c). Consider L(x)=f(x)-f(c)/(x-c) when x≠c and L(x)=lim_{x \to c} f(x)-f(c)/(x-c)=L(c) when x=c. Let y ∈ (a, b). Case 1: If y≠c, L(y)=f(y)-f(c)/(y-c) and lim_{x \to y} (f(x)-f(c))=y-c. Then lim_{x \to y} (x-c)=y-c and y-c≠0 as y≠c. Thus, lim_{x \to y} L(x) = lim_{x \to y} f(x)-f(c)/(x-c)=L(y), making L continuous. Case 2: If y=c, lim_{x \to c} L(x) = lim_{x \to c} f(x)-f(c)/(x-c)=L(c), by definition of L(c) and L(x). Therefore, L is continuous. Now since we know L is cts, we can look at f(x) − f(c) = L(x)(x − c) and we know that lim_{x \to c} L(x)=L(c). We also know that as x → c, (x-c)→0. Now take the limit as x → c of f(x) − f(c) = L(x)(x − c). lim_{x \to c} f(x)-f(c)=lim_{x \to c} L(x)(x-c) lim_{x \to c} f(x) - lim_{x \to c} f(c) = (lim_{x \to c} L(x))*(lim_{x \to c} (x-c)) lim_{x \to c} f(x) - f(c) = L(c)*0 lim_{x \to c} f(x) = f(c). Thus f is continuous at c. Sorry if this is poorly written and typed, I struggle with writing proofs.

I'm currently stuck on Exercise 13(a), Chapter 4 of Stein & Shakarchi's Fourier Analysis. The fact that the kernel $H_t(x)$ is a good kernel, hence $u(x,t)\rightarrow f(x)$ at each point of continuity of $f$, is not easy to prove. This will be shown in the next chapter. However, one can prove directly that $H_t(x)$ is "peaked" at $x=0$ as $t\rightarrow > 0$ in the following sense: (a) Show that $\int_{-1/2}^{1/2}|H_t(x)|^2dx$ is of the order of magnitude of $t^{-1/2}$ as $t\rightarrow 0$. More precisely, prove that $t^{1/2}\int_{-1/2}^{1/2}|H_t(x)|^2dx$ converges to a non-zero limit as $t\rightarrow 0$. (b) Prove that $\int_{-1/2}^{1/2}x^2|H_t(x)|^2dx=O(t^{1/2})$ as $t\rightarrow 0$. [Hint: For (a) compare the sum $\sum_{n=-\infty}^\infty e^{-cn^2t}$ with the integral $\int_{-\infty}^{\infty} e^{-cx^2t}dx$ where $c>0$. For (b) use $x^2\leq C(\sin\pi x)^2$ for $-1/2\leq x\leq 1/2$, and apply the mean value theorem to $e^{-cx^2t}$. Following the hint, the Fourier coefficient of $H_t(x)$ is $e^{-4\pi^2n^2t}$, so by Parseval's Identity $$\int_{-1/2}^{1/2}|H(t)|^2dt=\sum_{n=-\infty}^\infty|e^{-4\pi^2n^2t}|^2=\sum_{n=-\infty}^\infty e^{-8\pi^2n^2t}.$$ Now $$\left|\sum_{n=-\infty}^\infty e^{-8\pi^2n^2t}-\int_{-\infty}^\infty e^{-8\pi^2x^2t}dx\right|=\left|\sum_{n=-\infty}^\infty\int_n^{n+1}\left(e^{-8\pi^2n^2t}-e^{-8\pi^2x^2t}\right)dx\right|$$ which is supposed to be less than or equal to $$\sum_{n=-\infty}^\infty\left|e^{-8\pi^2\xi_n^2t}\cdot 16\pi^2\xi_nt\right|,\quad\xi_n\in[n,n+1]$$ according to a solution I've read, but I don't understand where this comes from. Any help would be greatly appreciated!

Two dimensional instantenious velocity field of an axisymetric turbulent free jet is provided in the data file Jet_0001.txt. The jet nozzel diameter is Djet = 0.01 m and the centerline of the jet is located at y0 = 0.025 m. The jet velocity at the nozzle is Ujet = 3 m/s. Velocity data is obtained using particle image velocimetry (PIV) technique in the central plane of the jet. The data file consists of four columns organized as follows x-coordinate (m) y-coordinate (m) u-velocity, uins (m/s) v-velocity, vins (m/s)     Note: The size of the array is 127 × 127 = 16129 Using MATLAB or any other software available a) Plot the instantenious velocity vector field in x – y plane. Normalize x, and y axis with Djet. Make sure that the jet center is at y0/Djet = 0 b) Plot the contours of the the streamwise velocity, uins /Ujet and show the colormap with appropriate labels. c) Where is the maximum uins located? Explain. d) Calculate the instantenious vorticity component z. For this purpose, develop a code to calculate z numerically. Use the finite diffenrece approximation to calculate the velocity gradients. Submit your code and comment on the regions of high/low vorticity. e) Normalize z with Djet and Ujet and plot the new dimensionless vorticity contour plot with axis x/Djet and y/Djet.