Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$
Show the Maclaurin Series for $ln(cos x)=-frac{1}{2}x^2-frac{1}{12}x^4 + ...$ (1st part of the question already worked out )
Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$
Given $x=frac{pi}{4}$
sequences-and-series taylor-expansion
edited Nov 20 at 12:09
Bernard
118k639112
asked Nov 20 at 12:08
Henias
615
add a comment |
Show the Maclaurin Series for $ln(cos x)=-frac{1}{2}x^2-frac{1}{12}x^4 + ...$ (1st part of the question already worked out )
Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$
Given $x=frac{pi}{4}$
sequences-and-series taylor-expansion
edited Nov 20 at 12:09
Bernard
118k639112
asked Nov 20 at 12:08
Henias
615
well $ln 2approx x$, to any $xinBbb C$
– Masacroso
Nov 20 at 12:11
Hint: $2=frac1{1/2}$
– Arthur
Nov 20 at 12:11
add a comment |
Show the Maclaurin Series for $ln(cos x)=-frac{1}{2}x^2-frac{1}{12}x^4 + ...$ (1st part of the question already worked out )
Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$
Given $x=frac{pi}{4}$
sequences-and-series taylor-expansion
edited Nov 20 at 12:09
Bernard
118k639112
asked Nov 20 at 12:08
Henias
615
Show the Maclaurin Series for $ln(cos x)=-frac{1}{2}x^2-frac{1}{12}x^4 + ...$ (1st part of the question already worked out )
Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$
Given $x=frac{pi}{4}$
sequences-and-series taylor-expansion
sequences-and-series taylor-expansion
edited Nov 20 at 12:09
Bernard
118k639112
asked Nov 20 at 12:08
Henias
615
edited Nov 20 at 12:09
Bernard
118k639112
asked Nov 20 at 12:08
Henias
615
edited Nov 20 at 12:09
Bernard
118k639112
edited Nov 20 at 12:09
Bernard
118k639112
edited Nov 20 at 12:09
Bernard
118k639112
118k639112
asked Nov 20 at 12:08
Henias
615
asked Nov 20 at 12:08
Henias
615
asked Nov 20 at 12:08
Henias
615
615
well $ln 2approx x$, to any $xinBbb C$
– Masacroso
Nov 20 at 12:11
Hint: $2=frac1{1/2}$
– Arthur
Nov 20 at 12:11
add a comment |
well $ln 2approx x$, to any $xinBbb C$
– Masacroso
Nov 20 at 12:11
Hint: $2=frac1{1/2}$
– Arthur
Nov 20 at 12:11
well $ln 2approx x$, to any $xinBbb C$
– Masacroso
Nov 20 at 12:11
well $ln 2approx x$, to any $xinBbb C$
– Masacroso
Nov 20 at 12:11
Hint: $2=frac1{1/2}$
– Arthur
Nov 20 at 12:11
Hint: $2=frac1{1/2}$
– Arthur
Nov 20 at 12:11
add a comment |
1 Answer
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Hint:
Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$
Edit for completion:
For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$
answered Nov 20 at 12:13
cansomeonehelpmeout
6,6433834
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$
Edit for completion:
For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$
answered Nov 20 at 12:13
cansomeonehelpmeout
6,6433834
add a comment |
Hint:
Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$
Edit for completion:
For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$
answered Nov 20 at 12:13
cansomeonehelpmeout
6,6433834
add a comment |
Hint:
Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$
Edit for completion:
For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$
answered Nov 20 at 12:13
cansomeonehelpmeout
6,6433834
Hint:
Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$
Edit for completion:
For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$
answered Nov 20 at 12:13
cansomeonehelpmeout
6,6433834
edited Nov 21 at 16:58
answered Nov 20 at 12:13
cansomeonehelpmeout
6,6433834
answered Nov 20 at 12:13
cansomeonehelpmeout
6,6433834
answered Nov 20 at 12:13
cansomeonehelpmeout
6,6433834
6,6433834
add a comment |
add a comment |
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well $ln 2approx x$, to any $xinBbb C$
– Masacroso
Nov 20 at 12:11
Hint: $2=frac1{1/2}$
– Arthur
Nov 20 at 12:11