prove that 4ln((sqrt(2))/2)=-2ln(2)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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bereich 4

Solution

f(x)=4

Interval Notation

f(x)=4

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Solution steps

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(

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

)

Matrix Eigenwerte

Lass A eine Quadratmatrix sein, η einen Vektor und λ ein Skalar, das die Bedingung Aη=λη, erfüllt.
Dann ist λ der Eigenwert, der dem Eigenvektor λ von A zugordnet ist.

Die Eigenwerte von A sind die Wurzeln der charakteristischen Gleichung det(A−λ I)=0

det((

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

)−λ(

1	0	0	0
0	1	0	0
0	0	1	0
0	0	0	1

)): λ⁴−18λ³+40λ²−2λ−24

Löse λ⁴−18λ³+40λ²−2λ−24=0: λ≈ −0.65863…, λ≈ 1.11137…, λ≈ 2.12615…, λ≈ 15.42110…

Die Eigenwerte sind:

λ≈ −0.65863…, λ≈ 1.11137…, λ≈ 2.12615…, λ≈ 15.42110…

Solve instead

domäne 4

invers 4

interzeption 4

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

prove that 4ln((sqrt(2))/2)=-2ln(2)

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