Here is a persuasive advertisement for ghostwriting services for the topic of linear transformation Tg:
**Unlock the Secrets of Linear Transformation Tg with Our Expert Ghostwriting Services**
Are you struggling to understand the properties of linear transformation Tg? Do you need help with writing a paper on its linearity, one-to-one and onto properties, and its isometry?
Look no further! Our team of expert writers specializes in providing high-quality ghostwriting services for students in linear algebra and related fields. With our help, you can:
✍️ Get a well-researched and well-written paper that meets your academic requirements
✍️ Save time and effort by letting our experts handle the research and writing
✍️ Improve your understanding of linear transformation Tg and its properties
✍️ Achieve academic success with a high-quality paper that earns you top grades
Our ghostwriting services cover a wide range of topics, including:
* Linearity and non-linearity of Tg
* One-to-one and onto properties of Tg
* Isometry and its applications
* Matrix representations and eigenvalues
Our team of experts has a deep understanding of linear algebra and its applications, ensuring that you receive a high-quality paper that meets your academic needs.
**Why choose our ghostwriting services?**
* Fast turnaround time: Get your paper in as little as 24 hours
* High-quality writing: Our experts have a proven track record of producing high-quality papers
* Confidentiality: Your privacy is guaranteed, and we never share your information with anyone
* Affordable pricing: Get expert help at a price that fits your budget
Don’t let linear transformation Tg hold you back from achieving academic success. Contact us today to learn more about our ghostwriting services and let us help you unlock the secrets of linear algebra!
1. Clearly \(T_{g}\) is linear; it is one-to-one and onto. The last claim follows since \(T_{g}\) has an inverse, namely
\[T_{g^{-1}}\left(h\right)=h\circ g^{-1}\ \ \ \text{where}\ \ g^{-1}:[0,1] \rightarrow[a,b]\]
with \(g^{-1}\left(t\right)=a+t\left(b-a\right)\) for \(0\leq x\leq 1.\)2. \(T_{g}\) is an isometry because \(g\) is onto and
\[||T_{g}(f)||_{\infty}=\max_{x\in\left[a,b\right]}|f\left(g\left(x\right)\right)|= \max_{t\in g\left(\left[a,b\right]\right)}|f\left(t\right)|=\max_{t\in\left[0,1 \right]}|f\left(t\right)|=||f||_{\infty}.\]
3. \(T_{g}\) maps polynomials to polynomials.
If \(p\left(t\right)=\sum_{k=0}^{n}a_{k}t^{k}\) is a polynomial, then
\[T_{g}\left(p\left(t\right)\right)=p\left(g\left(t\right)\right)=\sum_{k=0}^{n} a_{k}\left(g\left(t\right)\right)^{k}=\sum_{k=0}^{n}a_{k}\left(\frac{t-a}{b-a} \right)^{k}\]
which is also a polynomial.
**Proposition 29**.: \(C\left[0,1\right]\) _is separable._
**Idea of the proof.** Let \(f\in C\left[0,1\right]\) and let \(\epsilon>0\). We first approximate \(f\) by a polygonal function shown in Figure 2.15.
Since \(f\in C\left[0,1\right],\ f\) is uniformly continuous, and we can find a sufficiently large \(n\) so that
\[\left|f\left(x\right)-f\left(y\right)\right|<\epsilon\quad\text{whenever} \ \ |x-y|<\frac{1}{n}.\]
Now define a polygonal function \(g\) as
\[g\left(\frac{k}{n}\right)=f\left(\frac{k}{n}\right)\text{ for }k=0,…,n\]
and \(g\) is linear on each interval \(\left(\frac{k}{n},\frac{k+1}{n}\right);\) moreover,
\[\left\|f-g\right\|_{\infty}=\max_{a\leq t\leq b}\left|\left(f-g\right)\left(t \right)\right|<\epsilon.\]
Next we change our approximating function to \(h.\) Let \(h\) be another polygonal function such that
Figure 2.15: Approximation of \(f\) by a polygonal function
1. \(h\) also has nodes at \(\frac{k}{n}\) for \(k=0,1,…,n.\)
2. \(h\left(\frac{k}{n}\right)\) is rational.
3. \(\left\|h\left(\frac{k}{n}\right)-g\left(\frac{k}{n}\right)\right\|_{\infty}<\epsilon\) for each \(k.\)
Then \(\left\|h-g\right\|_{\infty}<\epsilon\) which will imply \(\left\|f-h\right\|_{\infty}<2\epsilon\). And the set of all polygonal functions taking only rational values at the nodes \(\left(\frac{k}{n}\right)_{k=0}^{n}\) for some \(n\) is countable. More precisely, for each \(n\), let \(Q_{n}\) be the set of all polygonal functions that have nodes at \(x=\frac{k}{n},k=0,1,...,n\) and take only rational values at these points. To finish the proof, show that \(Q_{n}\) is a countable set and \(\bigcup Q_{n}\) is a countable dense set in \(C[0,1].\)
### Weierstrass Approximation Theorem
If a function \(f\left(x\right)\) is continuous on a closed and bounded interval \(\left[a,b\right],\) then for each \(\epsilon>0\) there exists a polynomial \(P\left(x\right)\) such that \(\left|f\left(x\right)-P\left(x\right)\right|<\epsilon\) for all \(x\in\left[a,b\right].\)
**Remark 51**.: 1. Repeating the above theorem in the notation we have been using, given \(f\in C\left[a,b\right]\) and \(\epsilon>0\), there is a polynomial \(P\left(x\right)\) such that \(\left\|f-P\right\|_{\infty}<\epsilon.\) Hence there is a sequence of polynomials \(\left(P_{n}\right)\) such that
\[P_{n}\rightrightarrows f\text{ on }\left[a,b\right].\]
2. The Weierstrass approximation theorem yields another proof that \(C\left[a,b\right]\) is separable, meaning if \(\mathcal{P}\) is the set of polynomials, then \(\overline{\mathcal{P}}=\)\(C\left[a,b\right]\) where \(\overline{\mathcal{P}}\) is the closure of \(P\).
3. The theorem of Weierstrass is remarkable because continuous functions can have much more complicated behavior than polynomials. For example, recall Weierstrass’ construction of continuous functions that are not differentiable at any point. If a function \(f\) is sufficiently smooth, we can approximate it locally by partial sums of its Taylor series, but in general these polynomials do not provide the global approximation that the Weierstrass theorem provides.
### Bernstein Polynomials
**Definition 79**.: Let \(f\in C\left[0,1\right]\). The polynomial \(B_{n}(f)\) defined as
\[B_{n}\left(x,f\right)=\left(B_{n}\left(f\right)\right)\left(x\right)=\sum_{k=0 }^{n}f\left(\frac{k}{n}\right)\binom{n}{k}x^{k}\left(1-x\right)^{n-k},\text{ \ }0\leq x\leq 1\]
is called \(n\)th Bernstein polynomial of \(f.\)Note that \(B_{n}\left(f\right)\) is a polynomial of degree at most \(n\) and
\[\left(B_{n}\left(f\right)\right)\left(0\right)=f\left(0\right)\text{ and }\left(B_{n}\left(f\right)\right)\left(1\right)=f\left(1\right).\]
Later we will show that \(\left(B_{n}\left(f\right)\right)\left(x\right)\) is in some sense an “average” of the numbers \(f\left(\dfrac{k}{n}\right),k=0,…,n\).
**Example 85**.: Let \(f\left(x\right)=x^{2}\). To find \(\left(B_{1}\left(f\right)\right)\left(x\right)\) and \(\left(B_{2}\left(f\right)\right)\left(x\right)\) we set:
\[\left(B_{1}\left(f\right)\right)\left(x\right) = \sum_{k=0}^{1}f\left(k\right)\binom{1}{k}x^{k}\left(1-x\right)^{1-k}\] \[= f\left(0\right)\binom{1}{0}x^{0}\left(1-x\right)^{1}+f\left(1 \right)\binom{1}{1}x^{1}\left(1-x\right)^{0}\] \[= x.\]
\[\left(B_{2}\left(f\right)\right)\left(x\right) = \sum_{k=0}^{2}f\left(\dfrac{k}{2}\right)\binom{2}{k}x^{k}\left(1- x\right)^{2-k}\] \[= f\left(0\right)\binom{2}{0}x^{0}\left(1-x\right)^{2}+f\left( \dfrac{1}{2}\right)\binom{2}{1}x^{1}\left(1-x\right)^{1}\] \[+f\left(1\right)\binom{2}{2}x^{2}\left(1-x\right)^{0}\] \[= \frac{1}{4}\binom{2}{1}x\left(1-x\right)+\binom{2}{2}x^{2}\] \[= 2x\left(1-x\right)\frac{1}{4}+x^{2}=\frac{1}{2}x+\frac{1}{2}x^{ 2}.\]
In fact one can find
\[\left(B_{n}\left(f\right)\right)\left(x\right)=\frac{1}{n}x+\left(\frac{n-1}{ n}\right)x^{2}\]
and
\[\lim_{n\rightarrow\infty}B_{n}\left(f\right)=x^{2}.\]
**Remark 52**.:
* The binomial coefficient \(\binom{n}{k}=\dfrac{n!}{k!\left(n-k\right)!}\) and \(0!=1\) and the Binomial Theorem states that for \(0\leq k\leq n\), \[\left(x+y\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{k}y^{n-k}.\] Thus if we substitute \(y=1-x\) we obtain \[1=\sum_{k=0}^{n}\binom{n}{k}x^{k}\left(1-x\right)^{n-k}.\]
[MISSING_PAGE_EMPTY:542]
**Proof of the Bernstein’s Theorem.**
_Proof_. Observe that
\[\left|f\left(x\right)-\left(B_{n}\left(f\right)\right)\left(x\right)\right| = \left|f\left(x\right)-\sum_{k=0}^{n}f\left(\frac{k}{n}\right)r_{k} \left(x\right)\right|\] \[= \left|\sum_{k=0}^{n}\left(f\left(x\right)-f\left(\frac{k}{n} \right)\right)r_{k}\left(x\right)\right|\] \[\leq \sum_{k=0}^{n}\left|f\left(x\right)-f\left(\frac{k}{n}\right) \right|\left|r_{k}\left(x\right)\right|.\]
To prove that \(\left(B_{n}\left(f\right)\right)\left(x\right)\) tends uniformly to \(f\left(x\right)\) we appeal again to the uniform continuity of \(f\) on \(\left[0,1\right].\) Which means, for each \(\epsilon>0,\)\(\exists\delta>0\) such that \(\left|f\left(x\right)-f\left(y\right)\right|<\epsilon\) for all pairs of points \(x,y\in\left[0,1\right]\) with \(\left|x-y\right|<\delta.\) Now choose \(M\) such that \(\left|f\left(x\right)\right|\leq M\) on \(\left[0,1\right],\) and write
\[\left|f\left(x\right)-\left(B_{n}\left(f\right)\right)\left(x \right)\right| \leq \sum_{\left|k-nx\right|<\delta n}\left|f\left(x\right)-f\left( \frac{k}{n}\right)\right|\left|r_{k}\left(x\right)\right|\] \[+\sum_{\left|k-nx\right|\geq\delta n}\left|f\left(x\right)-f \left(\frac{k}{n}\right)\right|\left|r_{k}\left(x\right)\right|\] \[\leq S_{1}+S_{2}.\]
In the above equation \(S_{1}\) extends over all integers \(k\)\(\left(0\leq k\leq n\right)\) for which \(\left|x-\frac{k}{n}\right|<\delta\) and \(S_{2}\) extends over all integers \(k\) with \(\left|x-\frac{k}{n}\right|\geq\delta.\) By the uniform continuity of \(f\) and the fact that \(\sum_{k=0}^{n}r_{k}\left(x\right)=1\) when \(\left|k-nx\right|<\delta n\) we have \(\left|f\left(x\right)-f\left(\frac{k}{n}\right)\right|<\epsilon\) and thus
\[S_{1}\leq\sum_{k=0}^{n}\epsilon r_{k}\left(x\right)<\epsilon.\]
On the other hand, to get an upper bound for \(S_{2}\) we will use the boundedness of \(f\) on \(\left[0,1\right]\) to conclude
\[\left|f\left(x\right)-f\left(\frac{k}{n}\right)\right|\leq\left|f\left(x \right)\right|+\left|f\left(\frac{k}{n}\right)\right|\leq 2M,\]
发表回复