Question

बैसल असमिका कथन लिखकर सिद्ध कीजिए।
State and prove Bessel's inequality.
समद्विभाजन विधि से तीन पुनरावृति का उपयोग
धनात्मक मूल की गणना कीजिए।
Perform three iteration of the bisection
the equation f(x) = x2 - 5x + 2.
सिद्ध कीजिए कि किसी दिये गये क्षेत्र F पर दो
यदि और केवल यदि उनकी विमा समान ह
hat two finite dimensional
if their dimen​

Answer 1

Answer:

In mathematics, especially functional analysis, Bessel's inequality is a statement about the coefficients of an element {\displaystyle x}x in a Hilbert space with respect to an orthonormal sequence. The inequality was derived by F.W. Bessel in 1828.[1]

Let {\displaystyle H}H be a Hilbert space, and suppose that {\displaystyle e_{1},e_{2},...}e_1, e_2, ... is an orthonormal sequence in {\displaystyle H}H. Then, for any {\displaystyle x}x in {\displaystyle H}H one has

{\displaystyle \sum _{k=1}^{\infty }\left\vert \left\langle x,e_{k}\right\rangle \right\vert ^{2}\leq \left\Vert x\right\Vert ^{2},}{\displaystyle \sum _{k=1}^{\infty }\left\vert \left\langle x,e_{k}\right\rangle \right\vert ^{2}\leq \left\Vert x\right\Vert ^{2},}

where ⟨·,·⟩ denotes the inner product in the Hilbert space {\displaystyle H}H.[2][3][4] If we define the infinite sum

{\displaystyle x'=\sum _{k=1}^{\infty }\left\langle x,e_{k}\right\rangle e_{k},}x' = \sum_{k=1}^{\infty}\left\langle x,e_k\right\rangle e_k,

consisting of "infinite sum" of vector resolute {\displaystyle x}x in direction {\displaystyle e_{k}}e_{k}, Bessel's inequality tells us that this series converges. One can think of it that there exists {\displaystyle x'\in H}x' \in H that can be described in terms of potential basis {\displaystyle e_{1},e_{2},\dots }{\displaystyle e_{1},e_{2},\dots }.

For a complete orthonormal sequence (that is, for an orthonormal sequence that is a basis), we have Parseval's identity, which replaces the inequality with an equality (and consequently {\displaystyle x'}x' with {\displaystyle x}x).

Bessel's inequality follows from the identity

{\displaystyle {\begin{aligned}0\leq \left\|x-\sum _{k=1}^{n}\langle x,e_{k}\rangle e_{k}\right\|^{2}&=\|x\|^{2}-2\sum _{k=1}^{n}\operatorname {Re} \langle x,\langle x,e_{k}\rangle e_{k}\rangle +\sum _{k=1}^{n}|\langle x,e_{k}\rangle |^{2}\\&=\|x\|^{2}-2\sum _{k=1}^{n}|\langle x,e_{k}\rangle |^{2}+\sum _{k=1}^{n}|\langle x,e_{k}\rangle |^{2}\\&=\|x\|^{2}-\sum _{k=1}^{n}|\langle x,e_{k}\rangle |^{2},\end{aligned}}}{\displaystyle {\begin{aligned}0\leq \left\|x-\sum _{k=1}^{n}\langle x,e_{k}\rangle e_{k}\right\|^{2}&=\|x\|^{2}-2\sum _{k=1}^{n}\operatorname {Re} \langle x,\langle x,e_{k}\rangle e_{k}\rangle +\sum _{k=1}^{n}|\langle x,e_{k}\rangle |^{2}\\&=\|x\|^{2}-2\sum _{k=1}^{n}|\langle x,e_{k}\rangle |^{2}+\sum _{k=1}^{n}|\langle x,e_{k}\rangle |^{2}\\&=\|x\|^{2}-\sum _{k=1}^{n}|\langle x,e_{k}\rangle |^{2},\end{aligned}}}

which holds for any natural n.