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Question Number 99503 by DGmichael last updated on 21/Jun/20
Answered by MWSuSon last updated on 21/Jun/20
i think if you make sin^2 x=(1/2)(1−cos2x) it will be easier.. f(x)=sin^2 x f^n (x)=−(1/2)2^n cos(2x+((nπ)/2))=−2^(n−1) cos(2x+((nπ)/2))
$$\mathrm{i}\:\mathrm{think}\:\mathrm{if}\:\mathrm{you}\:\mathrm{make}\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos2x}\right) \\ $$$$\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\mathrm{easier}.. \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{f}^{\mathrm{n}} \left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2}^{\mathrm{n}} \mathrm{cos}\left(\mathrm{2x}+\frac{\mathrm{n}\pi}{\mathrm{2}}\right)=−\mathrm{2}^{\mathrm{n}−\mathrm{1}} \mathrm{cos}\left(\mathrm{2x}+\frac{\mathrm{n}\pi}{\mathrm{2}}\right) \\ $$
Commented by DGmichael last updated on 21/Jun/20
please I would like to know how to make them using dévelopment limited to order n .
Commented by MWSuSon last updated on 21/Jun/20
sorry sir I don't understand what you mean.
Commented by Ar Brandon last updated on 24/Jun/20
Using Taylor's series he meant. well I think so.
Answered by MWSuSon last updated on 21/Jun/20
you can make g(x) easier. g(x)=sinxcos3x=(1/2)[sin(4x)−sin(2x)]. ⇔ g(x)=(1/2)sin(4x)−(1/2)sin(2x) ⇒g^n (x)=(1/2)4^n sin(4x+((nπ)/2))−(1/2)2^n sin(2x+((nπ)/2)) ⇒g^n (x)=2^(2n−1) sin(4x+((nπ)/2))−2^(n−1) sin(2x+((nπ)/2))
$$\mathrm{you}\:\mathrm{can}\:\mathrm{make}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{easier}.\: \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{sinxcos3x}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}\left(\mathrm{4x}\right)−\mathrm{sin}\left(\mathrm{2x}\right)\right]. \\ $$$$\Leftrightarrow\:\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{4x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2x}\right) \\ $$$$\Rightarrow\mathrm{g}^{\mathrm{n}} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{4}^{\mathrm{n}} \mathrm{sin}\left(\mathrm{4x}+\frac{\mathrm{n}\pi}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2}^{\mathrm{n}} \mathrm{sin}\left(\mathrm{2x}+\frac{\mathrm{n}\pi}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{g}^{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{2}^{\mathrm{2n}−\mathrm{1}} \mathrm{sin}\left(\mathrm{4x}+\frac{\mathrm{n}\pi}{\mathrm{2}}\right)−\mathrm{2}^{\mathrm{n}−\mathrm{1}} \mathrm{sin}\left(\mathrm{2x}+\frac{\mathrm{n}\pi}{\mathrm{2}}\right) \\ $$
Answered by mathmax by abdo last updated on 21/Jun/20
we have f(x) =sin^2 x ⇒f(x) =((1−cos(2x))/2) ⇒ f^((n)) (x) =((1/2))^((n)) −(1/2)(cos(2x))^n so for n≥1 we get f^((n)) (x) =−(1/2)(((e^(2ix) +e^(−2ix) )/2))^((n)) =−(1/4)(e^(2ix) )^((n)) −(1/4)(e^(−2ix) )^((n)) =−(1/4){(2i)^n e^(2ix) +(−2i)^n e^(−2ix) } ⇒f^((2n)) (x) =−(1/4){ (2i)^(2n) e^(2ix) +(−2i)^(2n) e^(−2ix) } =−(4^n /4){ (−1)^n e^(2ix) +(−1)^n e^(−2ix) } =(−4)^(n−1) (2cos(2x)) =2(−4)^(n−1) cos(2x) f^((2n+1)) (x) =−(1/4){ (2i)^(2n+1) e^(2ix) +(−2i)^(2n+1) e^(−2ix) } =−(4^n /4) { 2i(−1)^n e^(2ix) −2i(−1)^n e^(−2ix) } =−2i×4^(n−1) {2isin(2x)} =4^n sin(2x)
$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\mathrm{n}\right)} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{2x}\right)\right)^{\mathrm{n}} \:\:\mathrm{so}\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{1}\:\:\mathrm{we}\:\mathrm{get}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{e}^{\mathrm{2ix}} +\mathrm{e}^{−\mathrm{2ix}} }{\mathrm{2}}\right)^{\left(\mathrm{n}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{e}^{\mathrm{2ix}} \right)^{\left(\mathrm{n}\right)} \:−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{e}^{−\mathrm{2ix}} \right)^{\left(\mathrm{n}\right)} \:=−\frac{\mathrm{1}}{\mathrm{4}}\left\{\left(\mathrm{2i}\right)^{\mathrm{n}} \:\mathrm{e}^{\mathrm{2ix}} \:\:+\left(−\mathrm{2i}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2ix}} \right\} \\ $$$$\Rightarrow\mathrm{f}^{\left(\mathrm{2n}\right)} \left(\mathrm{x}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\left(\mathrm{2i}\right)^{\mathrm{2n}} \:\mathrm{e}^{\mathrm{2ix}} \:+\left(−\mathrm{2i}\right)^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2ix}} \right\} \\ $$$$=−\frac{\mathrm{4}^{\mathrm{n}} }{\mathrm{4}}\left\{\:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{\mathrm{2ix}} \:+\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2ix}} \right\}\:=\left(−\mathrm{4}\right)^{\mathrm{n}−\mathrm{1}} \:\left(\mathrm{2cos}\left(\mathrm{2x}\right)\right) \\ $$$$=\mathrm{2}\left(−\mathrm{4}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{cos}\left(\mathrm{2x}\right) \\ $$$$\mathrm{f}^{\left(\mathrm{2n}+\mathrm{1}\right)} \left(\mathrm{x}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\left(\mathrm{2i}\right)^{\mathrm{2n}+\mathrm{1}} \:\mathrm{e}^{\mathrm{2ix}} \:+\left(−\mathrm{2i}\right)^{\mathrm{2n}+\mathrm{1}} \:\mathrm{e}^{−\mathrm{2ix}} \right\} \\ $$$$=−\frac{\mathrm{4}^{\mathrm{n}} }{\mathrm{4}}\:\left\{\:\mathrm{2i}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{\mathrm{2ix}} \:−\mathrm{2i}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2ix}} \right\} \\ $$$$=−\mathrm{2i}×\mathrm{4}^{\mathrm{n}−\mathrm{1}} \left\{\mathrm{2isin}\left(\mathrm{2x}\right)\right\}\:=\mathrm{4}^{\mathrm{n}} \:\mathrm{sin}\left(\mathrm{2x}\right) \\ $$
Answered by mathmax by abdo last updated on 21/Jun/20
another way for f(x) =sin^2 x ⇒f^′ (x) =2sinx cosx ⇒ for n>0 f^((n)) (x) =2(sinx cosx)^((n−1)) =_(leibniz) 2 Σ_(k=0) ^(n−1) C_(n−1) ^k (sinx)^((k)) (cosx)^((n−k)) =2 Σ_(k=0) ^(n−1) C_(n−1) ^k sin(x+((kπ)/2))cos(x+(((n−k)π)/2)) .
$$\mathrm{another}\:\mathrm{way}\:\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)\:=\mathrm{2sinx}\:\mathrm{cosx}\:\Rightarrow\:\mathrm{for}\:\mathrm{n}>\mathrm{0} \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\mathrm{2}\left(\mathrm{sinx}\:\mathrm{cosx}\right)^{\left(\mathrm{n}−\mathrm{1}\right)} \:\:=_{\mathrm{leibniz}} \:\:\:\:\mathrm{2}\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\:\mathrm{C}_{\mathrm{n}−\mathrm{1}} ^{\mathrm{k}} \:\left(\mathrm{sinx}\right)^{\left(\mathrm{k}\right)} \left(\mathrm{cosx}\right)^{\left(\mathrm{n}−\mathrm{k}\right)} \\ $$$$=\mathrm{2}\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\mathrm{C}_{\mathrm{n}−\mathrm{1}} ^{\mathrm{k}} \:\mathrm{sin}\left(\mathrm{x}+\frac{\mathrm{k}\pi}{\mathrm{2}}\right)\mathrm{cos}\left(\mathrm{x}+\frac{\left(\mathrm{n}−\mathrm{k}\right)\pi}{\mathrm{2}}\right)\:. \\ $$
Answered by mathmax by abdo last updated on 21/Jun/20
g(x) =sinx cos(3x) =cos((π/2)−x)cos(3x) =(1/2){cos(3x+(π/2)−x)+cos(3x−(π/2)+x)} =(1/2){ −sin(2x)+sin(4x)} =(1/2){ ((e^(i4x) −e^(−i4x) )/(2i))−((e^(i2x) −e^(−i2x) )/(2i))} =(1/(4i)){ e^(4ix) −e^(−4ix) +e^(−2ix) −e^(2ix) } ⇒ g^((n)) (x) =(1/(4i)){ 4i)^n e^(4ix) −(−4i)^n e^(−4ix) +(−2i)^n e^(−2ix) −(2i)^n e^(2ix) } and g^()n)) can be simplified...
$$\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{sinx}\:\mathrm{cos}\left(\mathrm{3x}\right)\:=\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)\mathrm{cos}\left(\mathrm{3x}\right)\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{cos}\left(\mathrm{3x}+\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)+\mathrm{cos}\left(\mathrm{3x}−\frac{\pi}{\mathrm{2}}+\mathrm{x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:−\mathrm{sin}\left(\mathrm{2x}\right)+\mathrm{sin}\left(\mathrm{4x}\right)\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{\mathrm{e}^{\mathrm{i4x}} −\mathrm{e}^{−\mathrm{i4x}} }{\mathrm{2i}}−\frac{\mathrm{e}^{\mathrm{i2x}} −\mathrm{e}^{−\mathrm{i2x}} }{\mathrm{2i}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4i}}\left\{\:\mathrm{e}^{\mathrm{4ix}} −\mathrm{e}^{−\mathrm{4ix}} +\mathrm{e}^{−\mathrm{2ix}} −\mathrm{e}^{\mathrm{2ix}} \right\}\:\Rightarrow \\ $$$$\left.\mathrm{g}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{4i}}\left\{\:\:\mathrm{4i}\right)^{\mathrm{n}} \:\mathrm{e}^{\mathrm{4ix}} −\left(−\mathrm{4i}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{4ix}} \:+\left(−\mathrm{2i}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2ix}} \:−\left(\mathrm{2i}\right)^{\mathrm{n}} \:\mathrm{e}^{\mathrm{2ix}} \right\} \\ $$$$\mathrm{and}\:\mathrm{g}^{\left.\right)\left.\mathrm{n}\right)} \:\mathrm{can}\:\mathrm{be}\:\mathrm{simplified}… \\ $$

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