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  • 세상의 모든 계산기 수학, 과학, 공학 이야기
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    • 푸리에 급수 테이블 , 푸리에 변환 테이블, Fourier Series Table, Fourier Transform Table

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      • 2024.11.15 - 00:05 2024.11.14 - 11:27 930

    출처 : https://ena.etsmtl.ca/pluginfile.php/137982/mod_resource/content/8/Fourier-table.pdf

     

    Fourier-table_1.png

    Fourier-table_2.png

     

     


     

    출처 : https://en.wikipedia.org/wiki/Fourier_transform

     

    Functional relationships, one-dimensional

     

    The Fourier transforms in this table may be found in [Erdélyi 1954] or [Kammler 2000, appendix].

      Function Fourier transform
    unitary, ordinary frequency
    Fourier transform
    unitary, angular frequency
    Fourier transform
    non-unitary, angular frequency
    Remarks
      \[f(x)\] \[\begin{align} &\widehat{f}(\xi) \triangleq \widehat {f_1}(\xi) \\&= \int_{-\infty}^\infty f(x) e^{-i 2\pi \xi x}\, dx \end{align}\] \[\begin{align} &\widehat{f}(\omega) \triangleq \widehat {f_2}(\omega) \\&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{-i \omega x}\, dx \end{align}\] \[\begin{align} &\widehat{f}(\omega) \triangleq \widehat {f_3}(\omega) \\&= \int_{-\infty}^\infty f(x) e^{-i \omega x}\, dx \end{align}\] Definitions
    101 \[a\, f(x) + b\, g(x)\] \[a\, \widehat{f}(\xi) + b\, \widehat{g}(\xi)\] \[a\, \widehat{f}(\omega) + b\, \widehat{g}(\omega)\] \[a\, \widehat{f}(\omega) + b\, \widehat{g}(\omega)\] Linearity
    102 \[f(x - a)\] \[e^{-i 2\pi \xi a} \widehat{f}(\xi)\] \[e^{- i a \omega} \widehat{f}(\omega)\] \[e^{- i a \omega} \widehat{f}(\omega)\] Shift in time domain
    103 \[f(x)e^{iax}\] \[\widehat{f} \left(\xi - \frac{a}{2\pi}\right)\] \[\widehat{f}(\omega - a)\] \[\widehat{f}(\omega - a)\] Shift in frequency domain, dual of 102
    104 \[f(a x)\] \[\frac{1}{|a|} \widehat{f}\left( \frac{\xi}{a} \right)\] \[\frac{1}{|a|} \widehat{f}\left( \frac{\omega}{a} \right)\] \[\frac{1}{|a|} \widehat{f}\left( \frac{\omega}{a} \right)\] Scaling in the time domain. If \[{{abs|a}}\] is large, then \[f(ax)\] is concentrated around 0 and
    \[\frac{1}{|a|}\hat{f} \left( \frac{\omega}{a} \right)\]
    spreads out and flattens.
    105 \[\widehat {f_n}(x)\] \[\widehat {f_1}(x) \ \stackrel{\mathcal{F}_1}{\longleftrightarrow}\ f(-\xi)\] \[\widehat {f_2}(x) \ \stackrel{\mathcal{F}_2}{\longleftrightarrow}\ f(-\omega)\] \[\widehat {f_3}(x) \ \stackrel{\mathcal{F}_3}{\longleftrightarrow}\ 2\pi f(-\omega)\] The same transform is applied twice, but x replaces the frequency variable (ξ or ω) after the first transform.
    106 \[\frac{d^n f(x)}{dx^n}\] \[(i 2\pi \xi)^n \widehat{f}(\xi)\] \[(i\omega)^n \widehat{f}(\omega)\] \[(i\omega)^n \widehat{f}(\omega)\] nth-order derivative.
    As \[f\] is a [[Schwartz space|Schwartz function]]
    106.5 \[\int_{-\infty}^{x} f(\tau) d \tau\] \[\frac{\widehat{f}(\xi)}{i 2 \pi \xi} + C \, \delta(\xi)\] \[\frac{\widehat{f} (\omega)}{i\omega} + \sqrt{2 \pi} C \delta(\omega)\] \[\frac{\widehat{f} (\omega)}{i\omega} + 2 \pi C \delta(\omega)\] Integration.[1] Note: \[\delta\] is the [[Dirac delta function]] and \[C\] is the average ([[DC component|DC]]) value of \[f(x)\] such that \[\int_{-\infty}^\infty (f(x) - C) \, dx = 0\]
    107 \[x^n f(x)\] \[\left (\frac{i}{2\pi}\right)^n \frac{d^n \widehat{f}(\xi)}{d\xi^n}\] \[i^n \frac{d^n \widehat{f}(\omega)}{d\omega^n}\] \[i^n \frac{d^n \widehat{f}(\omega)}{d\omega^n}\] This is the dual of 106
    108 \[(f * g)(x)\] \[\widehat{f}(\xi) \widehat{g}(\xi)\] \[\sqrt{2\pi}\ \widehat{f}(\omega) \widehat{g}(\omega)\] \[\widehat{f}(\omega) \widehat{g}(\omega)\] The notation \[f * g\] denotes the [[convolution]] of \[f\] and \[g\] — this rule is the [[convolution theorem]]
    109 \[f(x) g(x)\] \[\left(\widehat{f} * \widehat{g}\right)(\xi)\] \[ \frac{1}{\sqrt{2\pi}} \left(\widehat{f} * \widehat{g}\right)(\omega) \] \[\frac{1}{2\pi}\left(\widehat{f} * \widehat{g}\right)(\omega)\] This is the dual of 108
    110 For \[f(x)\] purely real \[\widehat{f}(-\xi) = \overline{\widehat{f}(\xi)}\] \[\widehat{f}(-\omega) = \overline{\widehat{f}(\omega)}\] \[\widehat{f}(-\omega) = \overline{\widehat{f}(\omega)}\] Hermitian symmetry. \[\overline{z}\] indicates the [[complex conjugate]].
    113 For \[f(x)\] purely imaginary \[\widehat{f}(-\xi) = -\overline{\widehat{f}(\xi)}\] \[\widehat{f}(-\omega) = -\overline{\widehat{f}(\omega)}\] \[\widehat{f}(-\omega) = -\overline{\widehat{f}(\omega)}\] \[\overline{z}\] indicates the [[complex conjugate]].
    114 \[\overline{f(x)}\] \[\overline{\widehat{f}(-\xi)}\] \[\overline{\widehat{f}(-\omega)}\] \[\overline{\widehat{f}(-\omega)}\] [[Complex conjugation]], generalization of 110 and 113
    115 \[f(x) \cos (a x)\] \[\frac{\widehat{f}\left(\xi - \frac{a}{2\pi}\right)+\widehat{f}\left(\xi+\frac{a}{2\pi}\right)}{2}\] \[\frac{\widehat{f}(\omega-a)+\widehat{f}(\omega+a)}{2}\] \[\frac{\widehat{f}(\omega-a)+\widehat{f}(\omega+a)}{2}\] This follows from rules 101 and 103 using [[Euler's formula]]:
    \[\cos(a x) = \frac{e^{i a x} + e^{-i a x}}{2}.\]
    116 \[f(x)\sin( ax)\] \[\frac{\widehat{f}\left(\xi-\frac{a}{2\pi}\right)-\widehat{f}\left(\xi+\frac{a}{2\pi}\right)}{2i}\] \[\frac{\widehat{f}(\omega-a)-\widehat{f}(\omega+a)}{2i}\] \[\frac{\widehat{f}(\omega-a)-\widehat{f}(\omega+a)}{2i}\] This follows from 101 and 103 using [[Euler's formula]]:
    \[\sin(a x) = \frac{e^{i a x} - e^{-i a x}}{2i}.\]

    [1] The Integration Property of the Fourier Transform

      Function Fourier transform
    unitary, ordinary frequency
    Fourier transform
    unitary, angular frequency
    Fourier transform
    non-unitary, angular frequency
    Remarks
      \[ f(x) \] \[ \begin{align} &\hat{f}(\xi) \triangleq \hat f_1(\xi) \\&= \int_{-\infty}^\infty f(x) e^{-i 2\pi \xi x}\, dx \end{align} \] \[ \begin{align} &\hat{f}(\omega) \triangleq \hat f_2(\omega) \\&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{-i \omega x}\, dx \end{align} \] \[ \begin{align} &\hat{f}(\omega) \triangleq \hat f_3(\omega) \\&= \int_{-\infty}^\infty f(x) e^{-i \omega x}\, dx \end{align} \] Definitions
    201 \[ \operatorname{rect}(a x) \] \[ \frac{1}{|a|}\, \operatorname{sinc}\left(\frac{\xi}{a}\right) \] \[ \frac{1}{\sqrt{2 \pi a^2}}\, \operatorname{sinc}\left(\frac{\omega}{2\pi a}\right) \] \[ \frac{1}{|a|}\, \operatorname{sinc}\left(\frac{\omega}{2\pi a}\right) \] The rectangular pulse and the normalized sinc function, here defined as \[ 1=\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \]
    202 \[ \operatorname{sinc}(a x) \] \[ \frac{1}{|a|}\, \operatorname{rect}\left(\frac{\xi}{a} \right) \] \[ \frac{1}{\sqrt{2\pi a^2}}\, \operatorname{rect}\left(\frac{\omega}{2 \pi a}\right) \] \[ \frac{1}{|a|}\, \operatorname{rect}\left(\frac{\omega}{2 \pi a}\right) \] Dual of rule 201. The rectangular function is an ideal low-pass filter, and the sinc function is the non-causal impulse response of such a filter. The sinc function is defined here as \[ 1=\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \]
    203 \[ \operatorname{sinc}^2 (a x) \] \[ \frac{1}{|a|}\, \operatorname{tri} \left( \frac{\xi}{a} \right) \] \[ \frac{1}{\sqrt{2\pi a^2}}\, \operatorname{tri} \left( \frac{\omega}{2\pi a} \right) \] \[ \frac{1}{|a|}\, \operatorname{tri} \left( \frac{\omega}{2\pi a} \right) \] The function \[ \text{tri}(x) \] is the triangular function
    204 \[ \operatorname{tri} (a x) \] \[ \frac{1}{|a|}\, \operatorname{sinc}^2 \left( \frac{\xi}{a} \right) \] \[ \frac{1}{\sqrt{2\pi a^2}} \, \operatorname{sinc}^2 \left( \frac{\omega}{2\pi a} \right) \] \[ \frac{1}{|a|} \, \operatorname{sinc}^2 \left( \frac{\omega}{2\pi a} \right) \] Dual of rule 203.
    205 \[ e^{- a x} u(x) \] \[ \frac{1}{a + i 2\pi \xi} \] \[ \frac{1}{\sqrt{2 \pi} (a + i \omega)} \] \[ \frac{1}{a + i \omega} \] The function \[ u(x) \] is the Heaviside unit step function and \[ a > 0 \].
    206 \[ e^{-\alpha x^2} \] \[ \sqrt{\frac{\pi}{\alpha}}\, e^{-\frac{(\pi \xi)^2}{\alpha}} \] \[ \frac{1}{\sqrt{2 \alpha}}\, e^{-\frac{\omega^2}{4 \alpha}} \] \[ \sqrt{\frac{\pi}{\alpha}}\, e^{-\frac{\omega^2}{4 \alpha}} \] This shows that, for the unitary Fourier transforms, the Gaussian function \[ e^{−αx^2} \] is its own Fourier transform for some choice of \[ α \]. For this to be integrable we must have \[ Re(α) > 0 \].
    208 \[ e^{-a|x|} \] \[ \frac{2 a}{a^2 + 4 \pi^2 \xi^2} \] \[ \sqrt{\frac{2}{\pi}} \, \frac{a}{a^2 + \omega^2} \] \[ \frac{2a}{a^2 + \omega^{2}} \] For \[ Re(a) > 0 \]. That is, the Fourier transform of a two-sided decaying exponential function is a Lorentzian function.
    209 \[ \operatorname{sech}(a x) \] \[ \frac{\pi}{a} \operatorname{sech} \left( \frac{\pi^2}{ a} \xi \right) \] \[ \frac{1}{a}\sqrt{\frac{\pi}{2}} \operatorname{sech}\left( \frac{\pi}{2 a} \omega \right) \] \[ \frac{\pi}{a}\operatorname{sech}\left( \frac{\pi}{2 a} \omega \right) \] Hyperbolic secant is its own Fourier transform
    210 \[ e^{-\frac{a^2 x^2}2} H_n(a x) \] \[ \frac{\sqrt{2\pi}(-i)^n}{a} e^{-\frac{2\pi^2\xi^2}{a^2}} H_n\left(\frac{2\pi\xi}a\right) \] \[ \frac{(-i)^n}{a} e^{-\frac{\omega^2}{2 a^2}} H_n\left(\frac \omega a\right) \] \[ \frac{(-i)^n \sqrt{2\pi}}{a} e^{-\frac{\omega^2}{2 a^2}} H_n\left(\frac \omega a \right) \] \[ H_n \] is the \[ n \]th-order Hermite polynomial. If \[ a = 1 \] then the Gauss–Hermite functions are eigenfunctions of the Fourier transform operator. For a derivation, see Hermite polynomial. The formula reduces to 206 for \[ n = 0 \].
      Function Fourier transform
    unitary, ordinary frequency
    Fourier transform
    unitary, angular frequency
    Fourier transform
    non-unitary, angular frequency
    Remarks
      \[ f(x) \] \[ \begin{align} &\hat{f}(\xi) \triangleq \hat f_1(\xi) \\&= \int_{-\infty}^\infty f(x) e^{-i 2\pi \xi x}\, dx \end{align} \] \[ \begin{align} &\hat{f}(\omega) \triangleq \hat f_2(\omega) \\&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{-i \omega x}\, dx \end{align} \] \[ \begin{align} &\hat{f}(\omega) \triangleq \hat f_3(\omega) \\&= \int_{-\infty}^\infty f(x) e^{-i \omega x}\, dx \end{align} \] Definitions
    301 \[ 1 \] \[ \delta(\xi) \] \[ \sqrt{2\pi}\, \delta(\omega) \] \[ 2\pi\delta(\omega) \] The distribution \[ ''δ''(''ξ'') \] denotes the [[Dirac delta function]].
    302 \[ \delta(x) \] \[ 1 \] \[ \frac{1}{\sqrt{2\pi}} \] \[ 1 \] Dual of rule 301.
    303 \[ e^{i a x} \] \[ \delta\left(\xi - \frac{a}{2\pi}\right) \] \[ \sqrt{2 \pi}\, \delta(\omega - a) \] \[ 2 \pi\delta(\omega - a) \] This follows from 103 and 301.
    304 \[ \cos (a x) \] \[ \frac{\delta\left(\xi - \frac{a}{2\pi}\right)+\delta\left(\xi+\frac{a}{2\pi}\right)}{2} \] \[ \sqrt{2 \pi}\,\frac{\delta(\omega-a)+\delta(\omega+a)}{2} \] \[ \pi\left(\delta(\omega-a)+\delta(\omega+a)\right) \] This follows from rules 101 and 303 using [[Euler's formula]]:
    \[ \cos(a x) = \frac{e^{i a x} + e^{-i a x}}{2}. \]
    305 \[ \sin( ax) \] \[ \frac{\delta\left(\xi-\frac{a}{2\pi}\right)-\delta\left(\xi+\frac{a}{2\pi}\right)}{2i} \] \[ \sqrt{2 \pi}\,\frac{\delta(\omega-a)-\delta(\omega+a)}{2i} \] \[ -i\pi\bigl(\delta(\omega-a)-\delta(\omega+a)\bigr) \] This follows from 101 and 303 using
    \[ \sin(a x) = \frac{e^{i a x} - e^{-i a x}}{2i}. \]
    306 \[ \cos \left( a x^2 \right) \] \[ \sqrt{\frac{\pi}{a}} \cos \left( \frac{\pi^2 \xi^2}{a} - \frac{\pi}{4} \right) \] \[ \frac{1}{\sqrt{2 a}} \cos \left( \frac{\omega^2}{4 a} - \frac{\pi}{4} \right) \] \[ \sqrt{\frac{\pi}{a}} \cos \left( \frac{\omega^2}{4a} - \frac{\pi}{4} \right) \] This follows from 101 and 207 using
    \[ \cos(a x^2) = \frac{e^{i a x^2} + e^{-i a x^2}}{2}. \]
    307 \[ \sin \left( a x^2 \right) \] \[ - \sqrt{\frac{\pi}{a}} \sin \left( \frac{\pi^2 \xi^2}{a} - \frac{\pi}{4} \right) \] \[ \frac{-1}{\sqrt{2 a}} \sin \left( \frac{\omega^2}{4 a} - \frac{\pi}{4} \right) \] \[ -\sqrt{\frac{\pi}{a}}\sin \left( \frac{\omega^2}{4a} - \frac{\pi}{4} \right) \] This follows from 101 and 207 using
    \[ \sin(a x^2) = \frac{e^{i a x^2} - e^{-i a x^2}}{2i}. \]
    308 \[ e^{-\pi i\alpha x^2} \] \[ \frac{1}{\sqrt{\alpha}}\, e^{-i\frac{\pi}{4}} e^{i\frac{\pi \xi^2}{\alpha}} \] \[ \frac{1}{\sqrt{2\pi \alpha}}\, e^{-i\frac{\pi}{4}} e^{i\frac{\omega^2}{4\pi \alpha}} \] \[ \frac{1}{\sqrt{\alpha}}\, e^{-i\frac{\pi}{4}} e^{i\frac{\omega^2}{4\pi \alpha}} \] Here it is assumed \[ \alpha \] is real. For the case that alpha is complex see table entry 206 above.
    309 \[ x^n \] \[ \left(\frac{i}{2\pi}\right)^n \delta^{(n)} (\xi) \] \[ i^n \sqrt{2\pi} \delta^{(n)} (\omega) \] \[ 2\pi i^n\delta^{(n)} (\omega) \] Here, \[ n \] is a [[natural number]] and \[ ''δ''{{isup|(''n'')}}\left(\xi\right) \] is the \[ n \]th distribution derivative of the Dirac delta function. This rule follows from rules 107 and 301. Combining this rule with 101, we can transform all [[polynomial]]s.
    310 \[ \delta^{(n)}(x) \] \[ (i 2\pi \xi)^n \] \[ \frac{(i\omega)^n}{\sqrt{2\pi}} \] \[ (i\omega)^n \] Dual of rule 309. \[ ''δ''{{isup|(''n'')}}\left(\xi\right) \] is the \[ n \]th distribution derivative of the Dirac delta function. This rule follows from 106 and 302.
    311 \[ \frac{1}{x} \] \[ -i\pi\sgn(\xi) \] \[ -i\sqrt{\frac{\pi}{2}}\sgn(\omega) \] \[ -i\pi\sgn(\omega) \] Here \[ sgn(\xi) \] is the [[sign function]]. Note that \[ \frac{1}{x} \] is not a distribution. It is necessary to use the [[Cauchy principal value]] when testing against [[Schwartz functions]]. This rule is useful in studying the [[Hilbert transform]].
    312 \[ \begin{align} &\frac{1}{x^n} \\ &:= \frac{(-1)^{n-1}}{(n-1)!}\frac{d^n}{dx^n}\log |x| \end{align} \] \[ -i\pi \frac{(-i 2\pi \xi)^{n-1}}{(n-1)!} \sgn(\xi) \] \[ -i\sqrt{\frac{\pi}{2}}\, \frac{(-i\omega)^{n-1}}{(n-1)!}\sgn(\omega) \] \[ -i\pi \frac{(-i\omega)^{n-1}}{(n-1)!}\sgn(\omega) \] \[ \frac{1}{x^n} \] is the [[homogeneous distribution]] defined by the distributional derivative
    \[ \frac{(-1)^{n-1}}{(n-1)!}\frac{d^n}{dx^n}\log|x| \]
    313 \[ |x|^\alpha \] \[ -\frac{2\sin\left(\frac{\pi\alpha}{2}\right)\Gamma(\alpha+1)}{|2\pi\xi|^{\alpha+1}} \] \[ \frac{-2}{\sqrt{2\pi}}\, \frac{\sin\left(\frac{\pi\alpha}{2}\right)\Gamma(\alpha+1)}{|\omega|^{\alpha+1}} \] \[ -\frac{2\sin\left(\frac{\pi\alpha}{2}\right)\Gamma(\alpha+1)}{|\omega|^{\alpha+1}} \] This formula is valid for \[ 0 > \alpha > -1 \]. For \[ \alpha > 0 \] some singular terms arise at the origin that can be found by differentiating 320. If \[ Re \alpha > -1 \], then \[ |x|^\alpha \] is a locally integrable function, and so a tempered distribution. The function \[ \alpha \mapsto |x|^\alpha \] is a holomorphic function from the right half-plane to the space of tempered distributions. It admits a unique meromorphic extension to a tempered distribution, also denoted \[ |x|^\alpha \] for \[ \alpha \neq -1, -3, ... \] (See [[homogeneous distribution]].)
      \[ \frac{1}{\sqrt{|x|}} \] \[ \frac{1}{\sqrt{|\xi|}} \] \[ \frac{1}{\sqrt{|\omega|}} \] \[ \frac{\sqrt{2\pi}}{\sqrt{|\omega|}} \] Special case of 313.
    314 \[ \sgn(x) \] \[ \frac{1}{i\pi \xi} \] \[ \sqrt{\frac{2}{\pi}} \frac{1}{i\omega} \] \[ \frac{2}{i\omega} \] The dual of rule 311. This time the Fourier transforms need to be considered as a [[Cauchy principal value]].
    315 \[ u(x) \] \[ \frac{1}{2}\left(\frac{1}{i \pi \xi} + \delta(\xi)\right) \] \[ \sqrt{\frac{\pi}{2}} \left( \frac{1}{i \pi \omega} + \delta(\omega)\right) \] \[ \pi\left( \frac{1}{i \pi \omega} + \delta(\omega)\right) \] The function \[ u(x) \] is the Heaviside [[Heaviside step function|unit step function]]; this follows from rules 101, 301, and 314.
    316 \[ \sum_{n=-\infty}^{\infty} \delta (x - n T) \] \[ \frac{1}{T} \sum_{k=-\infty}^{\infty} \delta \left( \xi -\frac{k }{T}\right) \] \[ \frac{\sqrt{2\pi }}{T}\sum_{k=-\infty}^{\infty} \delta \left( \omega -\frac{2\pi k}{T}\right) \] \[ \frac{2\pi}{T}\sum_{k=-\infty}^{\infty} \delta \left( \omega -\frac{2\pi k}{T}\right) \] This function is known as the [[Dirac comb]] function. This result can be derived from 302 and 102, together with the fact that
    \[ \begin{align} & \sum_{n=-\infty}^{\infty} e^{inx} \\ = {}& 2\pi\sum_{k=-\infty}^{\infty} \delta(x+2\pi k) \end{align} \] as distributions.
    317 \[ J_0 (x) \] \[ \frac{2\, \operatorname{rect}(\pi\xi)}{\sqrt{1 - 4 \pi^2 \xi^2}} \] \[ \sqrt{\frac{2}{\pi}} \, \frac{\operatorname{rect}\left(\frac{\omega}{2}\right)}{\sqrt{1 - \omega^2}} \] \[ \frac{2\,\operatorname{rect}\left(\frac{\omega}{2}\right)}{\sqrt{1 - \omega^2}} \] The function \[ J_0(x) \] is the zeroth order [[Bessel function]] of first kind.
    Attached file
    Fourier-table.pdf 56.0KB 41Fourier-table_1.png 353.7KB 39Fourier-table_2.png 351.6KB 46
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    아 그렇네요. 감사합니다. ^^ 2026 04.28 정적분 구간에 미지수가 있고, solve 를 사용할 수 없을 때 그 값을 확인하려면? https://allcalc.org/57087 `SOLVE` 기능 내에 `∫(적분)` 기호를 사용할 수 없을 때 뉴튼-랩슨법을 직접 사용하는 방법 2026 04.15 뉴턴-랩슨 적분 방정식 시각화 v1.0 body { font-family: 'Pretendard', -apple-system, BlinkMacSystemFont, "Segoe UI", Roboto, Helvetica, Arial, sans-serif; display: flex; flex-direction: column; align-items: center; background: #f8f9fa; padding: 40px 20px; margin: 0; color: #333; } .container { background: white; padding: 40px; border-radius: 20px; box-shadow: 0 15px 35px rgba(0,0,0,0.08); max-width: 900px; width: 100%; } header { border-bottom: 2px solid #f1f3f4; margin-bottom: 30px; padding-bottom: 20px; } h1 { color: #1a73e8; margin: 0 0 10px 0; font-size: 1.8em; } p.subtitle { color: #5f6368; margin: 0; font-size: 1.1em; } .equation-box { background: #f1f3f4; padding: 15px; border-radius: 10px; text-align: center; margin-bottom: 30px; font-size: 1.3em; } canvas { border: 1px solid #e0e0e0; border-radius: 12px; background: #fff; width: 100%; height: auto; display: block; } .controls { margin-top: 30px; display: flex; gap: 15px; align-items: center; justify-content: center; flex-wrap: wrap; } button { padding: 12px 25px; border: none; border-radius: 8px; background: #1a73e8; color: white; cursor: pointer; font-weight: 600; font-size: 1em; transition: all 0.2s; box-shadow: 0 2px 5px rgba(26,115,232,0.3); } button:hover { background: #1557b0; transform: translateY(-1px); box-shadow: 0 4px 8px rgba(26,115,232,0.4); } button:active { transform: translateY(0); } button.secondary { background: #5f6368; box-shadow: 0 2px 5px rgba(0,0,0,0.2); } button.secondary:hover { background: #4a4e52; } .status-badge { background: #e8f0fe; color: #1967d2; padding: 8px 15px; border-radius: 20px; font-weight: bold; font-size: 0.9em; } .explanation { margin-top: 40px; padding: 25px; background: #fff8e1; border-left: 5px solid #ffc107; border-radius: 8px; line-height: 1.8; } .explanation h3 { margin-top: 0; color: #856404; } .math-symbol { font-family: 'Times New Roman', serif; font-style: italic; font-weight: bold; color: #d93025; } .code-snippet { background: #202124; color: #e8eaed; padding: 2px 6px; border-radius: 4px; font-family: monospace; } 📊 Newton-Raphson 적분 방정식 시뮬레이터 미분적분학의 기본 정리(FTC)를 이용한 수치해석 시각화 목표 방정식: ∫₀ᴬ (2√x) dx = 20 을 만족하는 A를 찾아라! 계산 시작 (A 추적) 초기화 현재 반복: 0회 💡 시각적 동작 원리 (Newton-Raphson & FTC) Step 1 (오차 측정): 현재 A까지 쌓인 파란색 면적이 목표치(20)와 얼마나 차이나는지 계산합니다. Step 2 (FTC의 마법): 면적의 변화율(미분)은 그 지점의 그래프 높이 f(A)와 같습니다. Step 3 (보정): 다음 A = 현재 A - (면적 오차 / 현재 높이) 공식을 사용하여 A를 이동시킵니다. 결론: 오차를 현재 높이로 나누면, 오차를 메우기 위해 필요한 가로 길이(ΔA)가 나옵니다. 이 과정을 반복하면 정답에 도달합니다! const canvas = document.getElementById('graphCanvas'); const ctx = canvas.getContext('2d'); const iterText = document.getElementById('iterText'); // 수학 설정 const targetArea = 20; const f = (x) => Math.sqrt(x) * 2; // 피적분 함수 f(x) = 2√x const F = (x) => (4/3) * Math.pow(x, 1.5); // 정적분 결과 F(x) = ∫ 2√x dx = 4/3 * x^(3/2) let A = 1.5; // 초기값 let iteration = 0; let animating = false; // 그래프 드로잉 설정 const scale = 50; const offsetX = 60; const offsetY = 380; function drawGrid() { ctx.strokeStyle = '#f1f3f4'; ctx.lineWidth = 1; ctx.beginPath(); for(let i=0; i 2026 04.11 참값 : A = ±2√5 근사값 : A≈±4.472135954999579392818347   2026 04.10 fx-570 ES 입력 결과 초기값 입력   반복 수식 입력    반복 결과       2026 04.10
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