## Are these version of Cauchy’s Theorem (complex analysis) equivalent?

I am looking through different complex analysis texts and comparing their statement and proofs of Cauchy’s theorem. One version I found is: A second version of Cauchy’s Theorem is: Now I was wondering if the corollary of the second version is in fact equivalent to the first version. The biggest difference I see is that the first version makes it clear that we need f to be analytic inside the closed curve whereas the second version doesn’t say that. Is one of the four criterions equivalent to saying the interior of the curve is analytic?

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## Is this proof about Cauchy’s sequence correct?

Let $$(a_n)$$ be a sequence of real numbers and for each $$nin N$$, let $$f_n=a_1+a_2+a_3+…+a_n$$ and $$g_n$$=$$|a_1|+|a_2|+…+|a_n|$$. Prove that if $$(g_n)$$ is a Cauchy sequence, then so is $$(f_n)$$.

My attempt: Since $$(g_n)$$ is a Cauchy sequence, we have,

$$||a_1|+|a_2|+|a_3|+…|a_n|| for all $$ngeq m$$

This implies $$|a_1+a_2+…a_n| for all $$ngeq m$$

Thus $$(f_n)$$ is a Cauchy sequence.

## mg.metric geometry – Second case of induction in Cauchy’s arm lemma

What if in the second case of induction, vertices q1, q2 and qn are already collinear, angle αn > 90°, and α’n-1 is for example 90° . Then α*n-1 would not be larger than αn-1 and no edges will get stretched, thus |q1q*n| won’t be larger than |q1qn|. Did I miss something? Does Q have to be strictly convex, but Q’ only convex?

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## complex analysis – Prove an inequality by Cauchy’s Theorem

I am trying to solve the following question:

Let $$f(z)=c_0+c_1z+…+c_nz^n$$ be a polynomial，Prove that $$int_{-1}^1|f(x)|^2dxleqpiint_{0}^{2pi}|f(e^{itheta})|^2dfrac{dtheta}{2pi}$$

If all $$c_k$$ are real，it suffices to integral $$f(z)^2$$ along two semicircles,but I have no idea when $$c_k$$ are complex，can anyone help me?

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## real analysis – Prove Cauchy’s inequality: \$(sum_{i=1}^n a_i b_i)^2 le sum_{i=1}^n a_i^2 sum_{i=1}^n b_i^2.\$

Prove that for any real number $$a_i, b_i, i = 1,2,…,n$$:
$$(sum_{i=1}^n a_i b_i)^2 le sum_{i=1}^n a_i^2 sum_{i=1}^n b_i^2.$$
Proof in my book goes:
$$sum_{i=1}^n a_i^2 sum_{i=1}^n b_i^2-(sum_{i=1}^n a_i b_i)^2 = sum_{i,j=1}^na_i^2b_j^2-sum_{i,j=1}^na_ib_ia_jb_j =$$
$$=frac{1}{2}sum_{i,j=1}^n(a_ib_j-b_ia_j)^2ge0$$
But I can’t see why there is $$frac{1}{2}$$ and I would be glad if someone explained it.
Tia.

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## Complex analysis. Cauchy’s integral formula

I want to find $$f(1)$$ and $$f'(3)$$, where
$$f(z) = dfrac{1}{2pi i} int_{|zeta| = 2} dfrac {e^{1/({zeta – 5})}}{zeta – z} d zeta$$
where
the integral is taken over a positively oriented circle. I think, that $$f'(3) = 0$$ because of Cauchy theorem. But I have some problems with $$f(1)$$.

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## integration – Verification of an Cauchy’s contour Integral of Complementary Error function?

I tried to find an integral of the following,$$DeclareMathOperator{erfc}{erfc}$$

$$intlimits_0^{2pi} erfc(a + bcos(theta))erfc(c + dsin(theta)),dtheta$$

Where, $$a,b,c,d in Bbb R$$

Now, $$cos(theta) = frac{e^{jtheta} + e^{-jtheta}}{2};quadsin(theta) = frac{e^{jtheta } – e^{-jtheta }}{2j}$$

Let $$z = {e^{jtheta }}$$

Therefore, I can rewrite the integral as,
$$ointlimits_{|z| = 1} {erfcleft( {a + bleft( {frac{{z + {z^*}}}{2}} right)} right)erfcleft( {c + dleft( {frac{{z – {z^*}}}{{2j}}} right)} right)frac{{dz}}{{iz}}}$$

Again, let $$f(z) = erfcleft(a + bleft(frac{z + z^*}{2}right)right)erfcleft(c + dleft(frac{z – z^*}{2j}right) right)$$

Hence , the final integral can be written as,

$$ointlimits_{|z| = 1}f(z)frac{dz}{iz} = 2pi f(0) = 2pierfc(a)erfc(c)$$

As $$0$$ falls inside $$|z| = 1$$.

Can you tell me if I am correct or wrong or it needs more reasoning? Thank you.

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## ho.history overview – Did Euler ever use anything similar to Cauchy’s inequality?

This could be asked more provocatively, indeed how it arose, as “how did Euler do so much mathematics without using and/or knowing Cauchy’s inequality?”, something that came up in the context when reading a comment about how ubiquitous said inequality/variants is in various fields today.

But I think the title question is at least fact-based, and of historical interest.

Q. Did Euler ever use anything similar to Cauchy’s inequality?

Particularly I might have expected something like the integral version to appear somewhere in calculus of variations. Alternatively, something beyond simply a geometric-arithmetic mean might have occurred in number theory arguments. (I figure, as a first go, asking mathematicians is better than historians, especially as if there is “something similar” I’d like to hear it.)

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## cv.complex variables – Cauchy’s Integral with quadratic exponential term

As I was studying the Cauchy’s integral formula, I tried to do the integral:

$$begin{equation} I(x) = int_{-infty}^{infty} frac{1}{x – a} e^{(i A x^2 + i B x)} dx end{equation}$$
with $$A>0, B>0$$ and $$a > 0$$.

Consider an integral on a complex plan:
$$begin{equation} J(z) = int_{C + C_R} frac{1}{z – a} e^{(i A z^2 + i B z)} dz end{equation}$$
where $$C$$ is along the real axis $$-infty rightarrow +infty$$ and $$C_R$$ is the upper half circle $$z = Re^{itheta}$$ with $$R rightarrow infty$$ and $$theta in (0, 2pi)$$.

Naively, I would expect $$C_R$$ part of the integral gives zero and $$C$$ part of the integral gives $$I(x)$$, then the $$I(x)$$ can be derived by Cauchy’s integral formula.

However, as I tried to check the $$C_R$$ part of the integral, I found that:
$$begin{array} I_R(z = Re^{itheta}) &=& int_0^{pi} dtheta frac{iRe^{itheta}}{Re^{itheta} – a} exp(iAR^2e^{2itheta}+iBRe^{itheta}) end{array}$$

$$begin{array} |I_R| &leq & int_0^{pi} dtheta |frac{iRe^{itheta}}{Re^{itheta} – a}| |exp(iAR^2e^{2itheta}+iBRe^{itheta})| end{array}$$
where the first term

$$begin{equation} |frac{iRe^{itheta}}{Re^{itheta} – a}| leq frac{R}{R-a} rightarrow 1 as R rightarrow infty end{equation}$$

and the second term
$$begin{equation} |exp(iAR^2e^{2itheta}+iBRe^{itheta})| leq e^{-AR^2sin(2theta) – BRsin(theta)} end{equation}$$
will not approach to zero because of $$e^{-AR^2sin(2theta)}$$.

Is there anything wrong in my approach? And is there any other way I can perform this integral I(x)?