Search found 10 matches

by achilleas
Sat Jul 09, 2016 8:27 am
Forum: Real Analysis
Topic: Evaluate \(f^{\left (15 \right)}(0) \)
Replies: 1
Views: 1697

Re: Evaluate \(f^{\left (15 \right)}(0) \)

It's known that \(\sin x=\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\). for all \(x\in \mathbb{R}\). Hence \(\sin (x^3)=\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{3(2n+1)}}{(2n+1)!}=x^3-\dfrac{x^9}{3!}+\dfrac{x^{15}}{5!}-\cdots\). for all \(x\in \mathbb{R}\). ...
by achilleas
Thu Jul 07, 2016 5:27 pm
Forum: Calculus
Topic: integral inequality
Replies: 1
Views: 1839

Re: integral inequality

Let \(f(x)=\sin x-\dfrac{2x}{\pi}\) for \(x\in [0,\pi/2]\). Then \(f'(x)=\cos x-\dfrac{2}{\pi}\). Clearly, \( f'\) is strictly decreasing in \([0,\pi/2]\) and since \[ f'(\pi/2)>0>f'(0),\] there is a unique \(\xi\in (0,\pi/2)\) such that \(f'(\xi)=0\). Hence \(f'(x)>0\) for all \(x\in (\xi,\pi/2]\) ...
by achilleas
Thu Jul 07, 2016 5:22 pm
Forum: Calculus
Topic: How to draw tangents to an ellipse
Replies: 1
Views: 2014

Re: How to draw tangents to an ellipse

(a) It is easy to verify that the point \(P_b\) lies on the ellipse. Now, by implicit differentiation we get \[ \frac{2x}{4}+\frac{2y}{b^{2}}\frac{dy}{dx}=0 \] and so \[ \frac{dy}{dx}=-\frac{b^{2}}{4}\frac{x}{y}. \] Hence the tangent line at the ellipse at \(P_b=(1,\frac{\sqrt{3}b}{2})\) has equatio...
by achilleas
Thu Jul 07, 2016 5:22 pm
Forum: Calculus
Topic: How to draw tangents to an ellipse
Replies: 1
Views: 2014

How to draw tangents to an ellipse

Consider the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1\) (\(b>0\)). Let \(P_b=(1,\frac{\sqrt{3}b}{2})\) and let \(A\) be its projection on the x-axis, that is, \(A=(1,0)\). (a) Verify that the point \(P_b\) lies on the ellipse and find the equation of the tangent line \(\ell_b\) at the ellipse...
by achilleas
Thu Jul 07, 2016 4:49 pm
Forum: Calculus
Topic: Indefinite Integral
Replies: 2
Views: 2359

Re: Indefinite Integral

Integration by parts with \(u=(1-x)^{n-k}\) and \(dv=x^k dx\) gives us \(\int_0^1 x^k (1-x)^{n-k} dx= \frac{n-k}{k+1}\int_0^1 x^{k+1} (1-x)^{n-(k+1)} dx\), so \(\int_0^1 {n \choose k} x^k (1-x)^{n-k} dx= \int_0^1 {n \choose k+1} x^{k+1} (1-x)^{n-(k+1)} dx\), since \({n \choose k+1}=\frac{n-k}{k+1}{n...
by achilleas
Thu Jul 07, 2016 1:05 pm
Forum: Real Analysis
Topic: Convergence of Series
Replies: 1
Views: 2054

Convergence of Series

Let \(\alpha\) and \(\beta\) be real numbers, with \(\alpha>0\), and define the sequence \( \{a_n\}\) by \(a_1=\alpha\) and \(a_{n+1}=a_n\cdot \alpha^{a_n}\) for \(n=1,2,\ldots\). Study convergence of the series $$\sum_{n=1}^{\infty} \dfrac{a_n}{n^{\beta}}.$$
by achilleas
Thu Jul 07, 2016 12:27 pm
Forum: Calculus
Topic: Derivative of a Power Series
Replies: 2
Views: 2745

Derivative of a Power Series

Find the radius and interval of convergence of the power series
$$ x-\dfrac{x^3}{3\cdot 3!} + \dfrac{x^5}{5\cdot 5!}+...+(-1)^{n+1}\dfrac{x^{2n-1}}{(2n-1)\cdot(2n-1)!}+.... $$
and then evaluate \(f'\left(\frac{\pi}{2}\right)\), where \(f\) is the function defined by the above series.
by achilleas
Thu Jul 07, 2016 12:18 pm
Forum: Analysis
Topic: Non periodic function!
Replies: 3
Views: 3765

Re: Non periodic function!

After Demetres' answer, let us generalize this problem:

Find all polymomials \(p(x)\) such that \(\sin (p(x))\) is a periodic function.
by achilleas
Thu Jul 07, 2016 12:07 pm
Forum: Calculus
Topic: Three Collinear Points of Inflection
Replies: 1
Views: 1978

Three Collinear Points of Inflection

Show that the curve \(y=\frac{x+1}{x^{2}+1}\) has three points of inflection which lie on one straight line.
by achilleas
Mon Jan 18, 2016 3:46 am
Forum: Algebraic Structures
Topic: Group with p elements of order 2
Replies: 1
Views: 2121

Re: Group with p elements of order 2

Since \(G\) is non-abelian, \(p\) must be odd. Since \(p\) is a prime number that divides the order of \(G\), a theorem of Cauchy implies that \(G\) has an element of order \(p\), which generates a subgroup \(H\) of order \(p\). Since \(G\) is of order \(2p\), with \(p\) being odd, it has at most on...