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Symptoms

Technische Details Technische Daten anzeigen. CPU: Mediatek Helio G80 - 2 x 2GHz + 6 x 1,8 GHz RAM: 3 GB RAM oder 4 GB RAM Display: 2340 x 1080 - 6,5 Zoll Betriebssystem: Android 10, MIUI. Taylor Series A Taylor Series is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like x, x 2, x 3, etc.

You cannot connect to a server that is running Microsoft SQL Server 2014 or SQL Server 2012 by using the Transport Layer Security (TLS) protocol version 1.2. Additionally, the following error message is logged in the SQL Server error log:

TDSSNIClient initialization failed with error 0x80090331, status code 0x80. Reason: Unable to initialize SSL support. The client and server cannot communicate, because they do not possess a common algorithm.

Resolution

This problem was first fixed in the following cumulative update of SQL Server.

Cumulative Update 12 for SQL Server 2014
Cumulative Update 5 for SQL Server 2014 SP1
Cumulative Update 10 for SQL Server 2012 SP2
Cumulative Update 1 for SQL Server 2012 SP3Note This update adds the support for Transport Layer Security (TLS) protocol version 1.2 to SQL Server 2014 and to the Microsoft ODBC Driver for SQL Server.

About cumulative updates for SQL Server

Each new cumulative update for SQL Server contains all the hotfixes and all the security fixes that were included with the previous cumulative update. Check out the latest cumulative updates for SQL Server:

Note You can find information about the latest SQL Server builds from Where to find information about the latest SQL Server builds.

Status

Microsoft has confirmed that this is a problem in the Microsoft products that are listed in the 'Applies to' section.

References

Learn about the terminology Microsoft uses to describe software updates.

A Taylor Series is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like x, x², x³, etc.

Example: The Taylor Series for e^x

e^x = 1 + x + x²2! + x³3! + x⁴4! + x⁵5! + ...

(Note: ! is the Factorial Function.)

Does it really work? Let's try it:

Example: e^x for x=2

Well, we already know the answer is e² = 2.71828... × 2.71828... = 7.389056...

But let's try more and more terms of our infinte series:

It starts out really badly, but it then gets better and better!

Try using '2^n/fact(n)' and n=0 to 20 in the Sigma Calculator and see what you get.

Here are some common Taylor Series:

(There are many more.)

Approximations

We can use the first few terms of a Taylor Series to get an approximate value for a function.

Here we show better and better approximations for cos(x). The red line is cos(x), the blue is the approximation (try plotting it yourself) :

You can also see the Taylor Series in action at Euler's Formula for Complex Numbers.

What is this Magic?

Handy Note 1 0 9 X 2 1 2 In Strong Drive Sd Screws

How can we turn a function into a series of power terms like this?

Well, it isn't really magic. First we say we want to have this expansion:

f(x) = c₀ + c₁(x-a) + c₂(x-a)² + c₃(x-a)³ + ...

Then we choose a value 'a', and work out the values c₀ , c₁ , c₂ , ... etc

And it is done using derivatives (so we must know the derivative of our function)

Quick review: a derivative gives us the slope of a function at any point.

These basic derivative rules can help us:

• The derivative of a constant is 0
• The derivative of ax is a (example: the derivative of 2x is 2)
• The derivative of xⁿ is nx^n-1 (example: the derivative of x³ is 3x²)

We will use the little mark ' to mean 'derivative of'.

OK, let's start:

To get c₀, choose x=a so all the (x-a) terms become zero, leaving us with:

f(a) = c₀

So c₀ = f(a)

To get c₁, take the derivative of f(x):

f'(x) = c₁ + 2c₂(x-a) + 3c₃(x-a)² + ...

With x=a all the (x-a) terms become zero:

f'(a) = c₁

So c₁ = f'(a)

To get c₂, do the derivative again:

f''(x) = 2c₂ + 3×2×c₃(x-a) + ...

With x=a all the (x-a) terms become zero:

f''(a) = 2c₂

So c₂ = f''(a)/2

In fact, a pattern is emerging. Each term is

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• the next higher derivative ...
• ... divided by all the exponents so far multiplied together (for which we can use factorial notation, for example 3! = 3×2×1)

And we get:

f(x) = f(a) + f'(a)1!(x-a) + f'(a)2!(x-a)² + f''(a)3!(x-a)³ + ...

Now we have a way of finding our own Taylor Series:

For each term: take the next derivative, divide by n!, multiply by (x-a)ⁿ

Example: Taylor Series for cos(x)

f''(a) = 2c₂

So c₂ = f''(a)/2

In fact, a pattern is emerging. Each term is

1x0=

• the next higher derivative ...
• ... divided by all the exponents so far multiplied together (for which we can use factorial notation, for example 3! = 3×2×1)

And we get:

f(x) = f(a) + f'(a)1!(x-a) + f'(a)2!(x-a)² + f''(a)3!(x-a)³ + ...

Now we have a way of finding our own Taylor Series:

For each term: take the next derivative, divide by n!, multiply by (x-a)ⁿ

Example: Taylor Series for cos(x)

Start with:

f(x) = f(a) + f'(a)1!(x-a) + f'(a)2!(x-a)² + f''(a)3!(x-a)³ + ...

The derivative of cos is −sin, and the derivative of sin is cos, so:

• f(x) = cos(x)
• f'(x) = −sin(x)
• f'(x) = −cos(x)
• f''(x) = sin(x)
• etc...

And we get:

cos(x) = cos(a) − sin(a)1!(x-a) − cos(a)2!(x-a)² + sin(a)3!(x-a)³ + ...

Now put a=0, which is nice because cos(0)=1 and sin(0)=0:

cos(x) = 1 − 01!(x-0) − 12!(x-0)² + 03!(x-0)³ + 14!(x-0)⁴ + ...

Simplify:

cos(x) = 1 − x²/2! + x⁴/4! − ...

Try that for sin(x) yourself, it will help you to learn.

Or try it on another function of your choice.

The key thing is to know the derivatives of your function f(x).

Note: A Maclaurin Series is a Taylor Series where a=0, so all the examples we have been using so far can also be called Maclaurin Series.