Mixing
Showing an inequality of a complex exponential function
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If $T>2$ and exist $p$ such that $2(p-1)leq T<2p$ show that $int_{0}^{T}|e^{i npi t}|^2dxleq pint_{0}^{2}|e^{i n pi t}|^2dt.$
Hint: Use 2-periodicity of $e^{i npi t}$
I have this: $int_{0}^{T}|e^{i n pi t}|^2dtleq int_{0}^{2p} |e^{i n pi t}|^2dt=pint_{0}^{2}|e^{i npi sp}|^2ds$ with the change $s=t/p$
real-analysis
edited Jan 15 at 21:17
daw
24.5k1645
asked Jan 15 at 18:54
eraldcoileraldcoil
395211
$endgroup$
add a comment |
$begingroup$
If $T>2$ and exist $p$ such that $2(p-1)leq T<2p$ show that $int_{0}^{T}|e^{i npi t}|^2dxleq pint_{0}^{2}|e^{i n pi t}|^2dt.$
Hint: Use 2-periodicity of $e^{i npi t}$
I have this: $int_{0}^{T}|e^{i n pi t}|^2dtleq int_{0}^{2p} |e^{i n pi t}|^2dt=pint_{0}^{2}|e^{i npi sp}|^2ds$ with the change $s=t/p$
real-analysis
edited Jan 15 at 21:17
daw
24.5k1645
asked Jan 15 at 18:54
eraldcoileraldcoil
395211
$endgroup$
$begingroup$
Doesn't $int_0^2 e^{inpi t} , dt$ equal zero?
$endgroup$
– angryavian
Jan 15 at 19:04
$begingroup$
Sorry. Now, I fixed it.
$endgroup$
– eraldcoil
Jan 15 at 19:19
$begingroup$
This is just silly; $|e^{inpi t}|=1$.
$endgroup$
– David C. Ullrich
Jan 15 at 23:42
add a comment |
$begingroup$
If $T>2$ and exist $p$ such that $2(p-1)leq T<2p$ show that $int_{0}^{T}|e^{i npi t}|^2dxleq pint_{0}^{2}|e^{i n pi t}|^2dt.$
Hint: Use 2-periodicity of $e^{i npi t}$
I have this: $int_{0}^{T}|e^{i n pi t}|^2dtleq int_{0}^{2p} |e^{i n pi t}|^2dt=pint_{0}^{2}|e^{i npi sp}|^2ds$ with the change $s=t/p$
real-analysis
edited Jan 15 at 21:17
daw
24.5k1645
asked Jan 15 at 18:54
eraldcoileraldcoil
395211
$endgroup$
If $T>2$ and exist $p$ such that $2(p-1)leq T<2p$ show that $int_{0}^{T}|e^{i npi t}|^2dxleq pint_{0}^{2}|e^{i n pi t}|^2dt.$
Hint: Use 2-periodicity of $e^{i npi t}$
I have this: $int_{0}^{T}|e^{i n pi t}|^2dtleq int_{0}^{2p} |e^{i n pi t}|^2dt=pint_{0}^{2}|e^{i npi sp}|^2ds$ with the change $s=t/p$
real-analysis
real-analysis
edited Jan 15 at 21:17
daw
24.5k1645
asked Jan 15 at 18:54
eraldcoileraldcoil
395211
edited Jan 15 at 21:17
daw
24.5k1645
asked Jan 15 at 18:54
eraldcoileraldcoil
395211
edited Jan 15 at 21:17
daw
24.5k1645
edited Jan 15 at 21:17
daw
24.5k1645
edited Jan 15 at 21:17
daw
24.5k1645
24.5k1645
asked Jan 15 at 18:54
eraldcoileraldcoil
395211
asked Jan 15 at 18:54
eraldcoileraldcoil
395211
asked Jan 15 at 18:54
eraldcoileraldcoil
395211
395211
$begingroup$
Doesn't $int_0^2 e^{inpi t} , dt$ equal zero?
$endgroup$
– angryavian
Jan 15 at 19:04
$begingroup$
Sorry. Now, I fixed it.
$endgroup$
– eraldcoil
Jan 15 at 19:19
$begingroup$
This is just silly; $|e^{inpi t}|=1$.
$endgroup$
– David C. Ullrich
Jan 15 at 23:42
add a comment |
$begingroup$
Doesn't $int_0^2 e^{inpi t} , dt$ equal zero?
$endgroup$
– angryavian
Jan 15 at 19:04
$begingroup$
Sorry. Now, I fixed it.
$endgroup$
– eraldcoil
Jan 15 at 19:19
$begingroup$
This is just silly; $|e^{inpi t}|=1$.
$endgroup$
– David C. Ullrich
Jan 15 at 23:42
$begingroup$
Doesn't $int_0^2 e^{inpi t} , dt$ equal zero?
$endgroup$
– angryavian
Jan 15 at 19:04
$begingroup$
Doesn't $int_0^2 e^{inpi t} , dt$ equal zero?
$endgroup$
– angryavian
Jan 15 at 19:04
$begingroup$
Sorry. Now, I fixed it.
$endgroup$
– eraldcoil
Jan 15 at 19:19
$begingroup$
Sorry. Now, I fixed it.
$endgroup$
– eraldcoil
Jan 15 at 19:19
$begingroup$
This is just silly; $|e^{inpi t}|=1$.
$endgroup$
– David C. Ullrich
Jan 15 at 23:42
$begingroup$
This is just silly; $|e^{inpi t}|=1$.
$endgroup$
– David C. Ullrich
Jan 15 at 23:42
add a comment |
1 Answer
1
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$begingroup$
I have it!
$int_{0}^{2p} |e^{i npi it}|^2dt=int_{0}^{2} |e^{i npi it}|^2dt+cdots+ int_{2(p-1)}^{2p} |e^{i npi it}|^2dt=pint_{0}^{2p} |e^{i npi it}|^2dt$
answered Jan 15 at 19:33
eraldcoileraldcoil
395211
$endgroup$
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
$begingroup$
I have it!
$int_{0}^{2p} |e^{i npi it}|^2dt=int_{0}^{2} |e^{i npi it}|^2dt+cdots+ int_{2(p-1)}^{2p} |e^{i npi it}|^2dt=pint_{0}^{2p} |e^{i npi it}|^2dt$
answered Jan 15 at 19:33
eraldcoileraldcoil
395211
$endgroup$
add a comment |
$begingroup$
I have it!
$int_{0}^{2p} |e^{i npi it}|^2dt=int_{0}^{2} |e^{i npi it}|^2dt+cdots+ int_{2(p-1)}^{2p} |e^{i npi it}|^2dt=pint_{0}^{2p} |e^{i npi it}|^2dt$
answered Jan 15 at 19:33
eraldcoileraldcoil
395211
$endgroup$
add a comment |
$begingroup$
I have it!
$int_{0}^{2p} |e^{i npi it}|^2dt=int_{0}^{2} |e^{i npi it}|^2dt+cdots+ int_{2(p-1)}^{2p} |e^{i npi it}|^2dt=pint_{0}^{2p} |e^{i npi it}|^2dt$
answered Jan 15 at 19:33
eraldcoileraldcoil
395211
$endgroup$
I have it!
$int_{0}^{2p} |e^{i npi it}|^2dt=int_{0}^{2} |e^{i npi it}|^2dt+cdots+ int_{2(p-1)}^{2p} |e^{i npi it}|^2dt=pint_{0}^{2p} |e^{i npi it}|^2dt$
answered Jan 15 at 19:33
eraldcoileraldcoil
395211
answered Jan 15 at 19:33
eraldcoileraldcoil
395211
answered Jan 15 at 19:33
eraldcoileraldcoil
395211
answered Jan 15 at 19:33
eraldcoileraldcoil
395211
395211
add a comment |
add a comment |
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$begingroup$
Doesn't $int_0^2 e^{inpi t} , dt$ equal zero?
$endgroup$
– angryavian
Jan 15 at 19:04
$begingroup$
Sorry. Now, I fixed it.
$endgroup$
– eraldcoil
Jan 15 at 19:19
$begingroup$
This is just silly; $|e^{inpi t}|=1$.
$endgroup$
– David C. Ullrich
Jan 15 at 23:42