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Agohofidov6 Agohofidov6 2022-01-15

Problem utuxx=et in R×(0,+)

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John Stewart John Stewart 2022-01-15

Show that 1nj=1(1pj)n)0 as n

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Margie Marx Margie Marx 2022-01-14

Analytic iff Holomorphic on open domainss of C

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Miguel Reynolds Miguel Reynolds 2022-01-14

How do I proof that this is a metric in R?
Ive

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lugreget9 lugreget9 2022-01-14

The proof that if f(x) is continuous at x=c, then 1f(x) is continuous at x=c
If f(x) is continuous at x=c, f(x)0 then 1f(x) is continuous at x=c.
My proof: for any ϵ>0, we need to show there exists δ>0, such that whenever |xc|<δ, we have |1f(x)1f(c)|<ϵ.
|1f(x)1f(c)|=|f(x)f(c)f(x)f(c)|
=|f(x)f(c)|1|f(x)f(c)|
=|f(x)f(c)|1|f(x)|1|f(c)|
Since f is continuous at c, thus |f(x)f(c)|<ϵ. I need to find an upper bound of |f(x)|.
Because |f(x)f(c)|<ϵ, we then have
|f(c)|=|f(c)f(x)+f(x)||f(c)f(x)|+|f(x)|.
<ϵ+|f(x)|
Thus, we find |f(x)|>|f(c)|ϵ. Equivalently, 1|f(x)|<1|f(c)|ϵ. (Not rigorous here, since |f(c)|ϵ could be negative, how to remedy?)
Finally, (ignore not rigorous part)
|1f(x)1f(c)|<ϵ|f(c)|ϵ1|f(c
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PEEWSRIGWETRYqx PEEWSRIGWETRYqx 2022-01-14

I am trying to do excercise 4.1. in Continuous Time Markov Processes: An Introduction by Thomas Ligett. There I have to provide some counterexample of a function in C({0,1}N). However I dont

The Math analysis problems represent one of the most challenging problems for college students because the use of analysis is never easy. The best thing you can do is to see the answers to the questions that you have. You can take these analysis examples and examine each line to see how these came about. If you are unsure about something, it is good to start with the problem and pose it in different words. It will help you to achieve success with the logical proceeding as you provide analysis for the changes, alterations, assumptions, or mathematical-statistical data being used.