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Question Number 215893 by hardmath last updated on 20/Jan/25

$$\mathrm{Find}:$$$$1)\underset{\mathbf{x}\to 0}{\lim }\frac{1-\cos 2\mathrm{x}}{{\mathrm{x}}^2}=?$$$$2)\underset{\mathbf{n}=1}{\sum ^{\mathbf{n}=\mathrm{\infty }}}\frac{{\mathrm{n}}^2}{3^{\mathbf{n}}}=?$$

Answered by mathmax last updated on 23/Jan/25

$$cos\left(2x\right)\sim 1-\frac{4x^2}{2}=1-2x^2\Rightarrow \frac{1-cos\left(2x\right)}{x^2}\sim \frac{2x^2}{x^2}\Rightarrow$$$${lim}_{x\to 0}\frac{1-cos\left(2x\right)}{x^2}=2$$$$2)letS\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{x}^{n}and\mid x\mid <1\Rightarrow S^{{}^{\prime }}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{nx}^{n-1}$$$$and{xS}^{{}^{\prime }}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{nx}^{n}bydetivation$$$$S^{{}^{\prime }}\left(x\right)+{xS}^{\left(2\right)}\left(x\right)=\mathrm{\Sigma }{n}^2{x}^{n-1}\Rightarrow$$$${xS}^{{}^{\prime }}\left(x\right)+x^2{S}^{\left(2\right)}\left(x\right)=\sum _{n⩾1}{n}^2{x}^n$$$$x=\frac{1}{3}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{3^n}=\frac{1}{3}{S}^{{}^{\prime }}\left(\frac{1}{3}\right)+\frac{1}{9}{S}^{\left(2\right)}\left(\frac{1}{3}\right)$$$$S\left(x\right)=\frac{1}{1-x}\Rightarrow S^{{}^{\prime }}\left(x\right)=\frac{1}{{(1-x)}^2}$$$$et{S}^{\left(2\right)}\left(x\right)=-2\frac{(-1)(1-x)}{{(1-x)}^4}=\frac{2}{{(1-x)}^3}$$$$\Rightarrow S^{{}^{\prime }}\left(\frac{1}{3}\right)=\frac{1}{{\left(\frac{2}{3}\right)}^2}=\frac{9}{4}$$$$S^{\left(2\right)}\left(\frac{1}{3}\right)=\frac{2}{{\left(\frac{2}{3}\right)}^3}=2\times \frac{27}{8}=\frac{27}{4}\Rightarrow$$$$S=\frac{1}{3}.\frac{9}{4}+\frac{1}{9}.\frac{27}{4}=\frac{3}{4}+\frac{3}{4}=\frac{6}{4}=\frac{3}{2}$$

Answered by AntonCWX last updated on 21/Jan/25

$$1)$$$$Bydifferentiatingnumeratoranddenominatortwotimesweget$$$$\underset{x\to 0}{\lim }\left(\frac{4cos\left(2x\right)}{2}\right)=\underset{x\to 0}{\lim }\left(2cos\left(2x\right)\right)=2$$$$$$

Commented by hardmath last updated on 21/Jan/25

$$1.\mathrm{Answer}:1$$

Commented by AntonCWX last updated on 21/Jan/25

$$Sinceyouhavetheanswer.$$$$I{won}^{\prime }tbothertosolvethesecondone.$$

Commented by mr W last updated on 21/Jan/25

$$tohardmath:$$$$youshouldshowwhytheanswerfor$$$$1)is1.$$$$ithinkthesolutionandresultfrom$$$$AntonCWXsiriscorrect.$$

Answered by mr W last updated on 21/Jan/25

$$1)$$$$\underset{x\to 0}{\lim }\frac{1-\cos 2x}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x}{x^2}$$$$=2\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^2$$$$=2\times 1^2=2✓$$$$$$$$2)$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^n=\frac{1}{1-x}-1$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^{n-1}=\frac{1}{{(1-x)}^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^n=\frac{x}{{(1-x)}^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^2{x}^{n-1}=\frac{1}{{(1-x)}^2}+\frac{2x}{{(1-x)}^3}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^2{x}^n=\frac{x}{{(1-x)}^2}+\frac{2x^2}{{(1-x)}^3}=\frac{x(1+x)}{{(1-x)}^3}$$$$setx=\frac{1}{3}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2}{3^n}=\frac{\frac{1}{3}\times \frac{4}{3}}{{\left(\frac{2}{3}\right)}^3}=\frac{3}{2}✓$$

Answered by MathematicalUser2357 last updated on 21/Jan/25

$$thefirstone:1$$$$thesecondone:\frac{3}{2}$$

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