Recent questions in Composite functions

Composite functions
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Mara Boyd
2023-02-05

Composite functions
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Mark Rosales
2022-11-18

Composite functions
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Martin Hart
2022-10-31

$\mathbf{r}=(x,y)=x\mathbf{i}+y\mathbf{j}$

$\Vert r\Vert =\sqrt{{x}^{2}+{y}^{2}}$

Let's assume that $\mathbf{r}\ne 0$

We now define $f(x,y)={r}^{m}$

What is the right expression for $\mathrm{\nabla}f$?

1. $m{r}^{m-1}\mathbf{r}$

2. $m{r}^{m-2}\mathbf{r}$

3. $m{r}^{m-1}$

4. $m{r}^{0.5m-1}\mathbf{r}$

Reasoning was that it should be number $$1$$:

the derivative of $f$ according to $x$ comes out as $m{r}^{m-1}(\frac{x}{\sqrt{{x}^{2}+{y}^{2}}})$, while the derivative of $f$ according to y comes out as $m{r}^{m-1}(\frac{y}{\sqrt{{x}^{2}+{y}^{2}}})$. $\sqrt{{x}^{2}+{y}^{2}}=\Vert r\Vert =1$, and therefore this is equal to $m{r}^{m-1}\mathbf{r}$.

Composite functions
Answered

adarascarlet80
2022-10-05

Composite functions
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trapskrumcu
2022-09-26

$$y=(\mathrm{sin}x{)}^{\sqrt{x}}.$$

Composite functions
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gaby131o
2022-09-26

Composite functions
Answered

malaana5k
2022-09-24

I'm a bit confused how there can be a function of $y(t)$ inside of the function definition for $y(x)$.

I took the example that $y(x)=({x}^{2},x)$ and $f(y,z)=(y+z,y-z)$

$$\Rightarrow f(y(x))=f({x}^{2},x)=({x}^{2}+x,{x}^{2}-x)$$

And now if we follow the definition of $y(x)$ we get:

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}f(y(t))dt$$

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}({t}^{2}+t,{t}^{2}-t)dt$$

$$\Rightarrow y(x)=y({x}_{0})+(\frac{1}{3}{t}^{3}+\frac{1}{2}{t}^{2},\frac{1}{3}{t}^{3}-\frac{1}{2}{t}^{2}){{\textstyle |}}_{{x}_{0}}^{x}$$

$$\Rightarrow y(x)=(\frac{1}{3}{x}^{3}+\frac{1}{2}{x}^{2}+{C}_{1},\frac{1}{3}{x}^{3}-\frac{1}{2}{x}^{2}+{C}_{2})$$

Where is mistake?

Composite functions
Answered

videosfapaturqz
2022-09-24

I tried Schwarz lemma so that $|f(z)|\le |z|$, and I tried to use Weierstrass $$M$$ test, but I don't know how ${f}_{n}$ is bounded. How to solve this problem?

Composite functions
Answered

Camila Brandt
2022-09-23

My try:

$u=5x-1$, so $\frac{du}{dx}=5$, thus $dx=\frac{du}{5}$

How to cancel out the $x$ in front?

Composite functions
Answered

unjulpild9b
2022-09-20

Function of the form

$$f({x}^{2}+{y}^{2})$$

How do I find the partial derivatives

$$\frac{\mathrm{\partial}f}{\mathrm{\partial}y},\frac{\mathrm{\partial}f}{\mathrm{\partial}x}$$

How $f({x}^{2}+{y}^{2})$ behaves. Assuming it should of the form

$$g(x,y)\cdot 2y\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}h(x,y)\cdot 2x$$

Composite functions
Answered

Liam Potter
2022-09-20

For example,

$$f(g(x))=f(g(0))+{f}^{\prime}(g(0))g(x)+\frac{{f}^{\u2033}(g(0))}{2}{g}^{2}(x)+\cdots $$

In my case, $g(x)={e}^{-{x}^{2}}(0\le x\le 1)$.

Composite functions
Answered

Aidyn Meza
2022-09-20

where $$C$$ is the value of the composite zeta function at $$2$$ and $$P$$ is the prime zeta function at $$2$$. My question is this. What will be the value of $$C$$?

Composite functions
Answered

vballa15ei
2022-09-14

$f(x)={\displaystyle \frac{x}{x-1}}$

$g(x)={\displaystyle \frac{1}{x}}$

$h(x)={x}^{2}-1$

Find $f\circ g\circ h$ and state its domain.

The answer the textbook states is that the domain is all real values of $x$, except $\pm 1$ and $\pm \sqrt{2}$.

However surely the domain excludes $0$ as well, since $g(0)$ is undefined.

Composite functions
Answered

Modelfino0g
2022-09-14

Proof: Assume $f(x)$ is continuous at a point $L$, and that $\underset{x\to a}{lim}g(x)=L$.

$\mathrm{\forall}\u03f5>0,\mathrm{\exists}\delta >0:[|x-L|<\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\textstyle |}f(x)-f(L){\textstyle |}<\epsilon ]$.

And $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|g(x)-L<\delta ]$.

So, $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(g(x))-f(L)|<\u03f5]$.

$\underset{x\to a}{lim}g(x)=L$ so $f(\underset{x\to a}{lim}g(x))=f(L)$.

Composite functions
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spremani0r
2022-09-13

I have $f(x{)}^{\prime}=\frac{1}{(x+\sqrt{{a}^{2}+{x}^{2}})}\cdot (1+\frac{2x}{2\sqrt{{a}^{2}+{x}^{2}}})$, but can't simplify.

I want get $\frac{1}{\sqrt{{a}^{2}+{x}^{2}}}$

Composite functions
Answered

Spactapsula2l
2022-09-12

$$\begin{array}{rl}\frac{d}{dx}[\mathrm{sec}\left(\frac{x}{12}\right)]\text{}& =\mathrm{sec}\left(\frac{x}{12}\right)\ast \mathrm{tan}\left(\frac{x}{12}\right)\ast \frac{d}{dx}\left(\frac{x}{12}\right)\\ & =\mathrm{sec}\left(\frac{x}{12}\right)\ast \mathrm{tan}\left(\frac{x}{12}\right)\ast \frac{1}{12}\end{array}$$

If chain rule is not applied to this function like this because the function is "composite" which is why it should be done as the first variant, then how was chain rule altered for this function in the first variant?

Composite functions
Answered

Beckett Henry
2022-09-12

Composite functions
Answered

nar6jetaime86
2022-09-12

Assuming that f and g have reverse, ${f}^{-1}=h$ and ${g}^{-1}=s$ with $h:B\to A$, $s:C\to B$.

from that above i infer that the inverse of $(g\circ f)$ is $(s\circ g):C\to A$ that is ${g}^{-1}\circ {f}^{-1}=(g\circ f{)}^{-1}$; Hence for proof of $(g\circ f{)}^{-1}={f}^{-1}\circ {g}^{-1}$, proceed as before, only swapping functions, right?

Composite functions
Answered

andg17o7
2022-09-11

I have computed the first few terms:

${f}_{1}(x)=\frac{2-x}{3-2x}$

${f}_{2}(x)=\frac{3-2x}{4-3x}$

${f}_{3}(x)=\frac{4-3x}{5-4x}$

${f}_{4}(x)=\frac{5-4x}{6-5x}$

Then,

${f}_{n}(x)=\frac{n+1-nx}{n+2-(n+1)x}$

How to prove this with mathematical induction as it has variable $x$ in it as well as the $n$.

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