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| {{Distinguish|Cauchy's integral theorem}}
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| In [[mathematics]], '''Cauchy's integral formula''', named after [[Augustin-Louis Cauchy]], is a central statement in [[complex analysis]]. It expresses the fact that a [[holomorphic function]] defined on a disk is completely determined by its values on the boundary of the disk, and it provides integral formulas for all derivatives of a holomorphic function. Cauchy's formula shows that, in complex analysis, "differentiation is equivalent to integration": complex differentiation, like integration, behaves well under [[uniform convergence|uniform limits]] – a result denied in [[real analysis]].
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| ==Theorem==
| | In regards to baking and cooking, you do not have to limit yourself to gluten free cookbooks merely because you've a wheat allergy or Celiac disease. Actually, most recipes are easily made without gluten, a protein found in wheat, rye and barley products. Here's the best way to convert a recipe to gluten free.<br><br>It wants lots of preparation to follow gluten free diet. The consistently used cereals like barley, wheat, and rye contain high content of gluten. Hence, it is also seen in many types of bread. Grains for example wild rice, corn, buckwheat, millet, amaranth, quinoa, teff, oats, soybeans, and sunflower seeds comprise are certainly gluten free.<br><br>You probably already understand it if you've got one of these allergies. Keep in mind; however, attempting a brand new food may trigger a response you'd no idea you would have.<br><br>There are other forms of tests that would be done as a followup. If the blood test proves positive, then you certainly can get an elective biopsy done on your own intestines. A little piece assessed for damaged villi and will be cut out. These villi that are damaged will show you that you do, really, have celiac disease.<br><br>Much more specifically, there are two main proteins in gluten, gliadins and glutenins. One unique gliadin, found in wheat, is what leads to celiac diet sensitivity.<br><br>I've tried a few, although I haven't personally tried all of the various cookies in the product line. Their gingersnaps are my total favorite. I eat them for the flavor and not only because they are free. They're crisp and flavored. I really prefer them to anything else available on the market. Yet, I enjoy the peanut butter, oatmeal, and pecan shortbread. My husband adores the fudge- dipped grahams and the vanilla wafers while my mother is anything else with chocolate and a supporter of the double fudge. The only one in the line that I'm not extremely fond of is [http://graphicbay.org/blogs/2542/14991/what-services-are-offered-at-acc dr. Simring Dentist] the chocolate chip. They taste wonderful, but I like mine a little more and crispy on the chewy side.<br><br>Second course of the Thanksgiving raw foods vegan meal - Nut butters, fruit, sesame seeds, nuts, red cabbage salad with chopped raw vegetables and tempeh on a bed of arugula or spinach.<br><br>What do I eat instead? I eat loads of fresh fruit, vegetables, eggs, dairy product in temperance, a small meat, and some nuts. Then and now I 've some brown rice or oatmeal. At times, I eat products that are spelt but honestly after several years with minimal wheat, I simply don't have much interest in grains. Feeling good is a great bonus with my healthier way of eating. If someone could wave a magic wand and take away my intolerance to wheat, I 'm not really sure I 'd need them to. |
| Suppose ''U'' is an [[open subset]] of the [[complex plane]] '''C''', ''f'' : ''U'' → '''C''' is a holomorphic function and the closed disk
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| ''D'' = { ''z'' : | ''z'' − ''z''<sub>0</sub>| ≤ ''r''} is completely contained in ''U''. Let <math>\gamma</math> be the circle forming the [[boundary (topology)|boundary]] of ''D''. Then for every ''a'' in the [[interior (topology)|interior]] of ''D'':
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| :<math>f(a) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\, dz </math>
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| where the [[contour integral]] is taken [[Curve orientation|counter-clockwise]].
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| The proof of this statement uses the [[Cauchy integral theorem]] and similarly only requires ''f'' to be [[complex differentiable]]. Since the reciprocal of the denominator of the integrand in Cauchy's integral formula can be expanded as a [[power series]] in the variable (''a'' − ''z''<sub>0</sub>), it follows that [[holomorphic functions are analytic]]. In particular ''f'' is actually infinitely differentiable, with
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| :<math>f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\, dz.</math>
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| This formula is sometimes referred to as '''Cauchy's differentiation formula'''.
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| The circle ''γ'' can be replaced by any closed [[rectifiable curve]] in ''U'' which has [[winding number]] one about ''a''. Moreover, as for the Cauchy integral theorem, it is sufficient to require that ''f'' be holomorphic in the open region enclosed by the path and continuous on its [[closure (topology)|closure]]. | |
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| == Proof sketch ==
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| By using the Cauchy integral theorem, one can show that the integral over ''C'' (or the closed rectifiable curve) is equal to the same integral taken over an arbitrarily small circle around ''a''. Since ''f''(''z'') is continuous, we can choose a circle small enough on which ''f''(''z'') is arbitrarily close to ''f''(''a''). On the other hand, the integral
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| :<math>\oint_C \frac{1}{z-a} \,dz = 2 \pi i,</math>
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| over any circle ''C'' centered at ''a''. This can be calculated directly via a parametrization ([[integration by substitution]]) <math> z(t) = a + \varepsilon e^{it} </math> where 0 ≤ ''t'' ≤ 2''π'' and ''ε'' is the radius of the circle.
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| Letting ''ε'' → 0 gives the desired estimate
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| : <math>\begin{align}
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| \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-a} \,dz - f(a) \right |
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| &= \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)-f(a)}{z-a} \,dz \right |\\[.5em]
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| &= \left | \frac{1}{2\pi i}\int_0^{2\pi}\left(\frac{f(z(t))-f(a)}{\varepsilon\cdot e^{i\cdot t}}\cdot\varepsilon\cdot e^{t\cdot i}i\right )\,dt\right |\\
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| &\leq \frac{1}{2 \pi} \int_0^{2\pi} \frac{ |f(z(t)) - f(a)| } {\varepsilon} \,\varepsilon\,dt\\[.5em]
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| &\leq \max_{|z-a|=\varepsilon}|f(z) - f(a)|
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| \xrightarrow[\varepsilon\to 0]{} 0.
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| \end{align}</math>
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| == Example ==
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| [[File:ComplexResiduesExample.png|thumb|300px|Surface of the real part of the function ''g''(''z'') = ''z''<sup>2</sup> / (''z''<sup>2</sup> + 2''z'' + 2) and its singularities, with the contours described in the text.]]
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| Consider the function
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| :<math>g(z)=\frac{z^2}{z^2+2z+2}</math>
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| and the contour described by |''z''| = 2, call it ''C''.
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| To find the integral of ''g''(''z'') around the contour, we need to know the singularities of ''g''(''z''). Observe that we can rewrite ''g'' as follows:
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| :<math>g(z)=\frac{z^2}{(z-z_1)(z-z_2)}</math>
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| where <math>z_1=-1+i,</math> <math>z_2=-1-i.</math>
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| Clearly the poles become evident, their [[absolute value|moduli]] are less than 2 and thus lie inside the contour and are subject to consideration by the formula. By the [[Cauchy-Goursat theorem]], we can express the integral around the contour as the sum of the integral around ''z''<sub>1</sub> and ''z''<sub>2</sub> where the contour is a small circle around each pole<!-- diagram works best -->. Call these contours ''C''<sub>1</sub> around ''z''<sub>1</sub> and ''C''<sub>2</sub> around ''z''<sub>2</sub>.
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| Now, around ''C''<sub>1</sub>, ''f'' is [[holomorphic function|analytic]] (since the contour does not contain the other singularity), and this allows us to write ''f'' in the form we require, namely:
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| :<math>f(z)=\frac{z^2}{z-z_2}</math>
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| and now
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| :<math>\oint_C \frac{f(z)}{z-a}\, dz=2\pi i\cdot f(a)</math>
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| <!-- blank line -->
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| :<math>
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| \oint_{C_1} \frac{\left(\frac{z^2}{z-z_2}\right)}{z-z_1}\,dz
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| =2\pi i\frac{z_1^2}{z_1-z_2}.
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| </math>
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| Doing likewise for the other contour:
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| :<math>f(z)=\frac{z^2}{z-z_1},</math>
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| <!-- extra blank line so that adjacent lines of "displayed" TeX won't look too cluttered -->
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| :<math>
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| \oint_{C_2} \frac{\left(\frac{z^2}{z-z_1}\right)}{z-z_2}\,dz
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| =2\pi i\frac{z_2^2}{z_2-z_1}.
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| </math> | |
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| The integral around the original contour ''C'' then is the sum of these two integrals:
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| :<math>\begin{align}
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| \oint_C \frac{z^2}{z^2+2z+2}\,dz
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| &{}= \oint_{C_1} \frac{\left(\frac{z^2}{z-z_2}\right)}{z-z_1}\,dz
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| + \oint_{C_2} \frac{\left(\frac{z^2}{z-z_1}\right)}{z-z_2}\,dz \\[.5em]
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| &{}= 2\pi i\left(\frac{z_1^2}{z_1-z_2}+\frac{z_2^2}{z_2-z_1}\right) \\[.5em]
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| &{}= 2\pi i(-2) \\[.3em]
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| &{}=-4\pi i.
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| \end{align}</math>
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| An elementary trick using [[partial fraction decomposition]]:
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| :<math>
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| \oint_C g(z)dz
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| =\oint_C \left(1-\frac{1}{z-z_1}-\frac{1}{z-z_2}\right)dz
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| =0-2\pi i-2\pi i
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| =-4\pi i
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| </math>
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| ==Consequences==
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| The integral formula has broad applications. First, it implies that a function which is holomorphic in an open set is in fact [[infinitely differentiable]] there. Furthermore, it is an [[analytic function]], meaning that it can be represented as a [[power series]]. The proof of this uses the [[dominated convergence theorem]] and the [[geometric series]] applied to
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| :<math>f(\zeta) = \frac{1}{2\pi i}\int_C \frac{f(z)}{z-\zeta}\,dz.</math>
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| The formula is also used to prove the [[residue theorem]], which is a result for [[meromorphic function]]s, and a related result, the [[argument principle]]. It is known from [[Morera's theorem]] that the uniform limit of holomorphic functions is holomorphic. This can also be deduced from Cauchy's integral formula: indeed the formula also holds in the limit and the integrand, and hence the integral, can be expanded as a power series. In addition the Cauchy formulas for the higher order derivatives show that all these derivatives also converge uniformly.
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| The analog of the Cauchy integral formula in real analysis is the [[Poisson integral formula]] for [[harmonic function]]s; many of the results for holomorphic functions carry over to this setting. No such results, however, are valid for more general classes of differentiable or real analytic functions. For instance, the existence of the first derivative of a real function need not imply the existence of higher order derivatives, nor in particular the analyticity of the function. Likewise, the uniform limit of a sequence of (real) differentiable functions may fail to be differentiable, or may be differentiable but with a derivative which is not the limit of the derivatives of the members of the sequence.
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| Another consequence is that if {{math|1=''f''(''z'') = ∑ ''a''<sub>''n''</sub> ''z''<sup>''n''</sup> }} is holomorphic in |''z''| < ''R'' and 0 < ''r'' < ''R'' then the coefficients {{math|''a''<sub>''n''</sub>}} satisfy '''Cauchy's inequality'''<ref>{{harvnb|Titchmarsh|1939|p=84}}</ref>
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| :<math>\displaystyle{|a_n|\le r^{-n} \sup_{|z|=r}|f(z)|.}</math>
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| ==Generalizations==
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| ===Smooth functions===
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| A version of Cauchy's integral formula is the Cauchy-[[Dimitrie Pompeiu|Pompeiu]] formula,<ref>[http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:36.0454.04] [http://archive.numdam.org/ARCHIVE/AFST/AFST_1905_2_7_3/AFST_1905_2_7_3_265_0/AFST_1905_2_7_3_265_0.pdf] D. Pompeiu, ''Sur la continuité des fonctions de variables complexes'', Annales de la faculté des sciences de Toulouse Sér. 2, 7 no. 3 (1905), p. 265–315</ref> and holds for [[smooth function]]s as well, as it is based on [[Stokes' theorem]]. Let ''D'' be a disc in '''C''' and suppose that ''f'' is a complex-valued [[continuously differentiable function|''C''<sup>1</sup>]] function on the [[closure (topology)|closure]] of ''D''. Then {{harv|Hörmander|1966|loc=Theorem 1.2.1}}
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| :<math>f(\zeta) = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z) dz}{z-\zeta} + \frac{1}{2\pi i}\iint_D \frac{\partial f}{\partial \bar{z}}(z) \frac{dz\wedge d\bar{z}}{z-\zeta}.</math>
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| One may use this representation formula to solve the inhomogeneous [[Cauchy–Riemann equations]] in ''D''. Indeed, if ''φ'' is a function in ''D'', then a particular solution ''f'' of the equation is a holomorphic function outside the support of ''μ''. Moreover, if in an open set ''D'',
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| :<math>d\mu = \frac{1}{2\pi i}\phi \, dz\wedge d\bar{z}</math>
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| for some ''φ'' ∈ ''C''<sup>''k''</sup>(''D'') (''k'' ≥ 1), then <math>f(\zeta,\bar{\zeta})</math> is also in ''C''<sup>''k''</sup>(''D'') and satisfies the equation
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| :<math>\frac{\partial f}{\partial\bar{z}} = \phi(z,\bar{z}).</math>
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| The first conclusion is, succinctly, that the [[convolution]] ''μ''∗''k''(''z'') of a compactly supported measure with the '''Cauchy kernel'''
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| :<math>k(z) = \operatorname{p.v.}\frac{1}{z}</math>
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| is a holomorphic function off the support of ''μ''. Here p.v. denotes the [[Cauchy principal value|principal value]]. The second conclusion asserts that the Cauchy kernel is a [[fundamental solution]] of the Cauchy–Riemann equations. Note that for smooth complex-valued functions ''f'' of compact support on '''C''' the generalized Cauchy integral formula simplifies to | |
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| :<math>f(\zeta) = \frac{1}{2\pi i}\iint \frac{\partial f}{\partial \bar{z}}\frac{dz\wedge d\bar{z}}{z-\zeta},</math>
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| and is a restatement of the fact that, considered as a [[distribution (mathematics)|distribution]], <math>(\pi z)^{-1}</math> is a [[fundamental solution]] of the [[Cauchy-Riemann operator]] <math>\partial/\partial\overline{z}</math>.<ref>{{harvnb|Hörmander|1983|p=63,81}}</ref> The generalized Cauchy integral formula can be deduced for any bounded open region ''X'' with C<sup>1</sup> boundary ∂''X'' from this result and the formula for the [[distributional derivative]] of the [[indicator function|characteristic function]] χ<sub>''X''</sub> of ''X'':
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| :<math> {\partial \chi_X\over \partial \overline z}= {i\over 2} \oint_{\partial X} dz,</math>
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| where the distribution on the right hand side denotes [[contour integration]] along ∂''X''.<ref>{{harvnb|Hörmander|1983|pp=62–63}}</ref>
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| ===Several variables===
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| In [[several complex variables]], the Cauchy integral formula can be generalized to [[polydisc]]s {{harv|Hörmander|1966|loc=Theorem 2.2.1}}. Let ''D'' be the polydisc given as the [[Cartesian product]] of ''n'' open discs ''D''<sub>1</sub>, ..., ''D''<sub>''n''</sub>:
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| :<math>D = \prod_{i=1}^n D_i.</math>
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| Suppose that ''f'' is a holomorphic function in ''D'' continuous on the closure of ''D''. Then
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| :<math>f(\zeta) = \frac{1}{(2\pi i)^n}\int\cdots\iint_{\partial D_1\times\dots\times\partial D_n} \frac{f(z_1,\dots,z_n)}{(z_1-\zeta_1)\dots(z_n-\zeta_n)}dz_1\dots dz_n</math>
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| where ''ζ''=(''ζ''<sub>1</sub>,...,''ζ''<sub>''n''</sub>) ∈ ''D''.
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| ===In real algebras===
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| The Cauchy integral formula is generalizable to real vector spaces of two or more dimensions. The insight into this property comes from [[geometric algebra]], where objects beyond scalars and vectors (such as planar bivectors and volumetric trivectors) are considered, and a proper generalization of [[Stokes theorem]].
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| Geometric calculus defines a derivative operator <math>\nabla = \hat e_i \partial_i</math> under its geometric product—that is, for a <math>k</math>-vector field <math>\psi(\vec r)</math>, the derivative <math>\nabla \psi</math> generally contains terms of grade <math>k+1</math> and <math>k-1</math>. For example, a vector field (<math>k=1</math>) generally has in its derivative a scalar part, the divergence (<math>k=0</math>), and a bivector part, the curl (<math>k=2</math>). This particular derivative operator has a [[Green's function]]:
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| :<math>G(\vec r, \vec r') = \frac{1}{S_n} \frac{\vec r - \vec r'}{|\vec r - \vec r'|^n}</math>
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| where <math>S_n</math> is the surface area of a unit ball in the space (that is, <math>S_2=2\pi</math>, the circumference of a circle with radius 1, and <math>S_3 = 4\pi</math>, the surface area of a sphere with radius 1). By definition of a Green's function, <math>\nabla G(\vec r, \vec r') = \delta(\vec r- \vec r')</math>. It is this useful property that can be used, in conjunction with the generalized Stokes theorem:
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| :<math>\oint_{\partial V} d\vec S \; f(\vec r) = \int_V d\vec V \; \nabla f(\vec r)</math>
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| where, for an <math>n</math>-dimensional vector space, <math>d\vec S</math> is an <math>(n-1)</math>-vector and <math>d\vec V</math> is an <math>n</math>-vector. The function <math>f(\vec r)</math> can, in principle, be composed of any combination of multivectors. The proof of Cauchy's integral theorem for higher dimensional spaces relies on the using the generalized Stokes theorem on the quantity <math>G(\vec r,\vec r') f(\vec r')</math> and use of the product rule:
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| :<math>\oint_{\partial V'} G(\vec r, \vec r')\; d\vec S' \; f(\vec r') = \int_V \left([\nabla' G(\vec r, \vec r')] f(\vec r') + G(\vec r, \vec r') \nabla' f(\vec r')\right) \; d\vec V</math>
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| when <math>\nabla \vec f = 0</math>, <math>f(\vec r)</math> is called a ''monogenic function'', the generalization of holomorphic functions to higher-dimensional spaces—indeed, it can be shown that the Cauchy–Riemann condition is just the two-dimensional expression of the monogenic condition. When that condition is met, the second term in the right-hand integral vanishes, leaving only
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| :<math>\oint_{\partial V'} G(\vec r, \vec r')\; d\vec S' \; f(\vec r') = \int_V [\nabla' G(\vec r, \vec r')] f(\vec r') = -\int_V \delta(\vec r - \vec r') f(\vec r') \; d\vec V =- i_n f(\vec r)</math>
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| where <math>i_n</math> is that algebra's unit <math>n</math>-vector, the [[pseudoscalar]]. The result is
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| :<math>f(\vec r) =- \frac{1}{i_n} \oint_{\partial V} G(\vec r, \vec r')\; d\vec S \; f(\vec r') = -\frac{1}{i_n} \oint_{\partial V} \frac{\vec r - \vec r'}{S_n |\vec r - \vec r'|^n} \; d\vec S \; f(\vec r')</math>
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| Thus, as in the two-dimensional (complex analysis) case, the value of an analytic (monogenic) function at a point can be found by an integral over the surface surrounding the point, and this is valid not only for scalar functions but vector and general multivector functions as well.
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| ==See also==
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| *[[Cauchy–Riemann equations]]
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| *[[Methods of contour integration]]
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| *[[Nachbin's theorem]]
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| *[[Morera's theorem]]
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| *[[Mittag-Leffler's theorem]]
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| *[[Green's function]] generalizes this idea to the non-linear setup
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| *[[Schwarz integral formula]]
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| *[[Parseval–Gutzmer formula]]
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| ==Notes==
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| {{reflist}}
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| ==References==
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| * {{citation|first=Lars|last=Ahlfors|authorlink=Lars Ahlfors|title=Complex analysis|publisher=McGraw Hill|edition=3rd|year=1979|isbn=978-0-07-000657-7}}.
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| * [http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:36.0454.04] [http://archive.numdam.org/ARCHIVE/AFST/AFST_1905_2_7_3/AFST_1905_2_7_3_265_0/AFST_1905_2_7_3_265_0.pdf] D. Pompeiu, ''Sur la continuité des fonctions de variables complexes'', Annales de la faculté des sciences de Toulouse Sér. 2, 7 no. 3 (1905), p. 265–315
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| *{{citation|first=E.C.|last=Titchmarsh|title=Theory of functions|publisher=[[Oxford University Press]]|year=1939|edition=2nd}}
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| * {{citation|first=Lars|last=Hörmander|authorlink=Lars Hörmander|title=An introduction to complex analysis in several variables|publisher=Van Nostrand|year=1966}}
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| * {{citation|first=Lars|last=Hörmander|authorlink=Lars Hörmander|title=The Analysis of Linear Partial Differential Operators I|year=1983|publisher=Springer|isbn=3-540-12104-8}}
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| * {{citation |last1=Doran|first1=Chris |last2=Lasenby |first2=Anthony |title=Geometric Algebra for Physicists | publisher=Cambridge University Press|year=2003 |isbn=978-0-521-71595-9 }}
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| == External links ==
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| * {{springer|title=Cauchy integral|id=p/c020890}}
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| * {{MathWorld | urlname= CauchyIntegralFormula | title= Cauchy Integral Formula }}
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| * [http://math.fullerton.edu/mathews/c2003/IntegralRepresentationMod.html Cauchy Integral Formula Module by John H. Mathews]
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| {{DEFAULTSORT:Cauchy's Integral Formula}}
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| [[Category:Complex analysis]]
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| [[Category:Theorems in complex analysis]]
| |
In regards to baking and cooking, you do not have to limit yourself to gluten free cookbooks merely because you've a wheat allergy or Celiac disease. Actually, most recipes are easily made without gluten, a protein found in wheat, rye and barley products. Here's the best way to convert a recipe to gluten free.
It wants lots of preparation to follow gluten free diet. The consistently used cereals like barley, wheat, and rye contain high content of gluten. Hence, it is also seen in many types of bread. Grains for example wild rice, corn, buckwheat, millet, amaranth, quinoa, teff, oats, soybeans, and sunflower seeds comprise are certainly gluten free.
You probably already understand it if you've got one of these allergies. Keep in mind; however, attempting a brand new food may trigger a response you'd no idea you would have.
There are other forms of tests that would be done as a followup. If the blood test proves positive, then you certainly can get an elective biopsy done on your own intestines. A little piece assessed for damaged villi and will be cut out. These villi that are damaged will show you that you do, really, have celiac disease.
Much more specifically, there are two main proteins in gluten, gliadins and glutenins. One unique gliadin, found in wheat, is what leads to celiac diet sensitivity.
I've tried a few, although I haven't personally tried all of the various cookies in the product line. Their gingersnaps are my total favorite. I eat them for the flavor and not only because they are free. They're crisp and flavored. I really prefer them to anything else available on the market. Yet, I enjoy the peanut butter, oatmeal, and pecan shortbread. My husband adores the fudge- dipped grahams and the vanilla wafers while my mother is anything else with chocolate and a supporter of the double fudge. The only one in the line that I'm not extremely fond of is dr. Simring Dentist the chocolate chip. They taste wonderful, but I like mine a little more and crispy on the chewy side.
Second course of the Thanksgiving raw foods vegan meal - Nut butters, fruit, sesame seeds, nuts, red cabbage salad with chopped raw vegetables and tempeh on a bed of arugula or spinach.
What do I eat instead? I eat loads of fresh fruit, vegetables, eggs, dairy product in temperance, a small meat, and some nuts. Then and now I 've some brown rice or oatmeal. At times, I eat products that are spelt but honestly after several years with minimal wheat, I simply don't have much interest in grains. Feeling good is a great bonus with my healthier way of eating. If someone could wave a magic wand and take away my intolerance to wheat, I 'm not really sure I 'd need them to.