Matematik
hjælp til ligning opgave
a) skal jeg bare løse ligningen for x når k=1? altså virkelig bare solve?
b) den er jeg dælme i tvivl om :(
Svar #1
20. september 2021 af janhaa
a)
k=1
https://www.wolframalpha.com/input/?i=sqrt%281-x%5E2%29+%3Dx*tan%28x%29
k= 1/4
https://www.wolframalpha.com/input/?i=sqrt%28%281%2F4%29%5E2-x%5E2%29+%3Dx*tan%28x%29
Svar #2
20. september 2021 af janhaa
b)
k = 10
https://www.wolframalpha.com/input/?i=plot+sqrt%2810-x%5E2%29+-x*tan%28x%29+%3D+0+from+0+to+100
k = 100
https://www.wolframalpha.com/input/?i=plot+sqrt%28100-x%5E2%29+-x*tan%28x%29+%3D+0+from+0+to+100
Svar #3
20. september 2021 af janhaa
b)
k = 1000
https://www.wolframalpha.com/input/?i=plot+sqrt%281000-x%5E2%29+-x*tan%28x%29+%3D+0+from+0+to+100
k = 10000
https://www.wolframalpha.com/input/?i=plot+sqrt%2810000-x%5E2%29+-x*tan%28x%29+%3D+0+from+0+to+100
Svar #6
20. september 2021 af Anders521
#5 A close cousin to the Newton method is the Secant method. At least here one doesn't require the derivative of a function.
Svar #7
20. september 2021 af janhaa
#5 But I would like to show it with Mathematics and not graph
https://en.wikipedia.org/wiki/Newton%27s_method
Newton's method is calculus math
Svar #8
20. september 2021 af janhaa
#6#5 A close cousin to the Newton method is the Secant method. At least here one doesn't require the derivative of a function.
true
Svar #9
21. september 2021 af KaspermedK
I mean.. Show question b with mathematics.. I cannot se how I should use thsee links for question b.
Svar #10
21. september 2021 af Anders521
#9
Perhaps for the argument one can use the fact that f(x) :=sqrt( k - x2) is a continuous function whereas g(x) := x·tan(x) is not, but rather a periodic discontinuous function. So as k increases in size the domain and codomain of f increases causing its graph to intersect with the graph of g many times over... Just a thought.
Svar #11
21. september 2021 af oppenede
Da x > 0 kan ligningen ved division med x omskrives til:
tan(x) - √(k/x2 - 1) = 0
tan(x) er voksende fra -∞ til ∞ på hvert af intervallerne (π/2, 3π/2), (3π/2, 5π/2), osv.
Da -√(k/x2 - 1) er begrænset på hvert interval vil summen også antage hele i hvert interval.
Der er en løsning i hvert interval jf. mellemværdisætningen.
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