All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 124483 by snipers237 last updated on 03/Dec/20
Developthefunctionf(x)=exsinxThenDeducethat∑[n−12]k=0(−1)kCn2k+1=2n2sin(nπ4)
Answered by mindispower last updated on 03/Dec/20
f(x)=exsin(x)=Im(ex(1+i))f(n)=Im(1+i)nex(1+i)=Im{2n2einπ4ex(1+i)}x=0=2n2sin(nπ4)(sin(x)ex)(n)=∑nk=0Cnksin(k)(x)(ex)(n−k)(a)istheathderivation(sin(x)ex)(n)=f(n)(x)=∑k⩽nCnksin(x+kπ2)ex∣x=0=sin(0+kπ2)=0,k=2m=(−1)m∣k=2m+1=∑[n−12]k=0sin((2k+1)π2)Cn2k+1=f(n)(0)=2n2sin(nπ4)⇒∑k⩽[n−12](−1)kCn2k+1=2n2sin(nπ4)
Answered by mathmax by abdo last updated on 03/Dec/20
f(x)=Im(ex+ix)=Im(e(1+i)x)bute(1+i)x=∑n=0∞(1+i)nxnn!=∑n=0∞1n!(2)neinπ4xn=∑n=0∞(2)nn!{cos(nπ4)+isin(nπ4)}xn⇒exsinx=∑n=0∞2n2n!sin(nπ4)xn⇒f(n)(0)=2n2sin(nπ4)f(n)(x)=∑k=0nCnk(sinx)(k)(ex)(n−k)=∑k=0nCnksin(x+kπ2)ex⇒f(n)(0)=∑k=0nCnksin(kπ2)=∑k=0[n−12]Cn2k+1sin((2k+1)π2)=∑k=0[n−12](−1)kCn2k+1⇒2n2sin(nπ4)=∑k=0[n−12](−1)kCn2k+1
Terms of Service
Privacy Policy
Contact: info@tinkutara.com