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Bas_tet [7]
3 years ago
12

I need help with this​

Mathematics
1 answer:
frutty [35]3 years ago
4 0

Answer:

A.

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
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