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3 years ago

I need help with this

Mathematics

1 answer:

frutty [35]3 years ago

4 0

Answer:

Step-by-step explanation:

Twust me n' dwont doubt me

A is for apple

And cause A looks like ABC

DWUHHH

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Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

3 years ago

If f(x)=2x, then f^-1(x)=

-BARSIC- [3]

$f(x)=2x\to y=2x\\\\\text{change x and y}\\\\x=2y\\\\\text{solve for y}\\\\2y=x\qquad|:2\\\\y=\dfrac{x}{2}\\\\Answer:\ f^{-1}(x)=\dfrac{x}{2}$

8 0

4 years ago

“If f(x) and it’s inverse function, f^-1(x), are both plotted on the same coordinate plane, where is their point of intersection

Ghella [55]

Answer:

it is not always possible to find their intersection point for e.g take

y=x-1 it's inverse is y=x+1 so their is no intersection point but generally u find intersection points by equalizing their functions

f(x)=f^-1(x)

7 0

4 years ago

Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately norm

Olegator [25]

Answer:

ssshscuodalsjkhfydarysrysgssystayrsgsyrsytdthsusjtdu dmhxhmxmhxfulddmhxhmxmhxfuldkcjdjydjyfiycjyfjfjyxkydkhdkyykdmhxhmxmhxfuldkcjdjydmhxhmxmhxfudmhx5eiysncmgztjdengd.vzmgsyjskydyk mhxjfjnjyfgdjgjgxddhhuyyyyyyyyyrgryidgzfhzhgxfhskgfnznhhtgdastuwrustute5ueutei your t7riyri tut if6if6if8tiyroufi6f86ti7fo7t68t7ot97to7r7or86r9r7ot8ti6r78rz. xgmzgstjstjsjydjytdgnjgjajgsajttjatuasjtyjssiygjsstutsiutssjtahtarhDHhrarwuekyeiyekyjgskgsgkstjsgjssjgsjggsjgjssykskgskykhddkhdkhddkhghhyhhgggggggtr or r f f f f f g f f f. if f eheur r rrurur r rjeeur eurhrr

4 0

3 years ago

How do you write 37 divided 3 distributive porperty

Eduardwww [97]

Answer:

37 divided by 3 = 12.333333333.....

You can’t write it in distributive property, that only applies to multiplication.

Step-by-step explanation:

4 0

3 years ago

I need help with this​

I need help with this