Solve 2*pi(x/2)^2*sin(x) | Microsoft Math Solver

Evaluate

\frac{π x ^{2} s i n ( x )}{2}

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Differentiate w.r.t. x

\frac{π x ( x c o s ( x ) + 2 s i n ( x ) )}{2}

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5 problems similar to:

2 \cdot π (\frac{x}{2})^{2} \cdot sin (x)

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Similar Problems from Web Search

x_{k} = 2^{k} \cdot sin \frac{π}{2 ^{k}}

x_{1}^{'} = 2, x_{2}^{'} = 22, x_{k + 1} = x_{k} \frac{2 x _{k}}{x _{k} + x _{k - 1}}

https://math.stackexchange.com/questions/528480/prove-that-x-k-2k-cdot-sin-frac-pi2k-equals-x-1-2-x-2

You can use that

cos (2 x) = cos^{2} x - sin^{2} x,

1 - cos \frac{π}{2 ^{k}} = 1 - cos^{2} \frac{π}{2 ^{k + 1}} + sin^{2} \frac{π}{2 ^{k + 1}} = 2 sin^{2} \frac{π}{2 ^{k + 1}} .

\frac{2}{1 + c o s \frac{π}{2 ^{k}}} = \frac{2 ( 1 - c o s \frac{π}{2 ^{k}} )}{1 - c o s ^{2} \frac{π}{2 ^{k}}} = \frac{4 s i n ^{2} \frac{π}{2 ^{k + 1}}}{s i n ^{2} \frac{π}{2 ^{k}}} = \frac{2 s i n \frac{π}{2 ^{k + 1}}}{s i n \frac{π}{2 ^{k}}},

Lipschitz-continuous

f (x) = x^{2} \cdot sin (\frac{1}{x})

https://math.stackexchange.com/questions/618438/lipschitz-continuous-fx-x2-cdot-sin-left-frac1x-right

f^{'} (x) = {2 x sin \frac{1}{x} - cos \frac{1}{x} 0 if x \neq = 0 if x = 0 .

∣ f^{'} (x) ∣ \leq 2 ∣ x ∣ \cdot ∣ ∣ ∣ sin \frac{1}{x} ∣ ∣ ∣ + ∣ ∣ ∣ cos \frac{1}{x} ∣ ∣ ∣ \leq 3 \forall x \neq = 0 .

x^{α} \cdot sin (1 / x)

is absolutely continuous on

(0, 1)

https://math.stackexchange.com/questions/294827/prove-that-x-alpha-cdot-sin1-x-is-absolutely-continuous-on-0-1

1 < α < 2

. If not, as remarked by the questioner,

f (x)

is Lipschitz on

(0, 1)

and the problem is simpler. You can use the following approach. Take a small

δ > 0

Fourier transform of signal

t sin^{2} (t) / (π t)^{2}

https://math.stackexchange.com/questions/1598695/fourier-transform-of-signal-t-sin2t-pi-t2

HINTS: From the Convolution Theorem , we have

\int_{- \infty}^{\infty} f (t) g (t) e^{i ω t} d t = \frac{1}{2 π} \int_{- \infty}^{\infty} F (ω - ω^{'}) G (ω^{'}) d ω^{'}

f (t) = g (t) = \frac{s i n ( t )}{π t}

Using the Fund. Theorem of Calc find

h (x) = \int_{2}^{\frac{1}{x}} sin^{4} t d t

https://math.stackexchange.com/questions/3033249/using-the-fund-theorem-of-calc-find-hx-int-2-frac1x-sin4tdts-deri

F^{'} (x) : = sin^{4} (x)

F

is the antiderivative of

sin^{4} (x)

. Then by the fundamental theorem of calculus

h (x) = F (\frac{1}{x}) - F (2) .

Looking for a function

f

n

-differentiable, but

f^{(n)}

is not continuous

https://math.stackexchange.com/q/58329

n

be a positive integer and let

f (x) = x^{2 n} \cdot sin (\frac{1}{x})

f (0) = 0

Then mathematical induction can be used to prove: (a) The

n

th derivative of

f (x)

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