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Dec
31
comment Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$
It was only the way I wrote things that seemed not formal enough to me. Looking at the answer to this post, it looks like that's only my perception that was wrong. I did not start with the definition of $\tan(x) = \frac{\sin(x)}{\cos(x)}$ but, if I am not mistaken, I think this definition is based on the exact thing that makes me uneasy in the way I wrote the first steps of the limit above
Dec
31
accepted Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$
Dec
31
awarded  Scholar
Dec
31
comment Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$
Looks indeed better to me. Isn't this definition based on the "opp/hyp/adj" definition though ?
Dec
30
asked Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$
Aug
8
awarded  Tumbleweed
May
19
awarded  Supporter
Apr
22
answered Finding ALL solutions to $2(\sin^2(x)) - 5\sin(x)-3 = 0$?
Apr
14
awarded  Teacher
Dec
15
comment Autocorrelation derivation using fourier transform
Oh my god yeah I was tired when I answered you. $R_{x}(\tau) = \frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}{cos(\omega{t})cos(\omega(t-\tau))dt‌​}$. Anyway, it doesn't change the result of the auto-correlation. However, My question is more about the fourier derivation of the auto-correlation, no matter how much I could be wrong/inexact with the integral derivation. I would be pleased if you could comment on that in priority !
Dec
15
comment Autocorrelation derivation using fourier transform
Obviously, for the derivation using the "integral" derivation, the following definition has been used : $R_{x} = \frac{1}{2\pi}\int_{0}^{2\pi}cos(\omega{t})cos(\omega(t-\tau))dt$ as the integral of a (co)sine is not defined over infinity. However, I do not get how it should affect the "fourier" derivation ?
Dec
14
asked Autocorrelation derivation using fourier transform
Aug
3
answered odd person out game
Jul
31
revised Solving $5^n > 4,000,000$ without a calculator
edited body
Jul
31
comment Solving $5^n > 4,000,000$ without a calculator
Oww ... I wanted to state that $10^7 > 4.10^6$ ... Thanks for pointing this out ! :>
Jul
31
comment Solving $5^n > 4,000,000$ without a calculator
may you explain the comment please ? Something bad in the answer ?
Jul
31
awarded  Editor
Jul
31
revised Solving $5^n > 4,000,000$ without a calculator
added 241 characters in body
Jul
31
revised Solving $5^n > 4,000,000$ without a calculator
latex correction
Jul
31
answered Solving $5^n > 4,000,000$ without a calculator