# Hello from a rusty old (84) maths enthusiast.

Discussion in 'Introductions' started by alan whittle, Dec 17, 2019.

1. ### alan whittle

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My name is Alan Whittle, and I hope to get some answers and give some answers. I graduated in maths in 1972, and retired from work in 2000, but still fiddle with maths in various ways, although more slowly than I used to.

alan whittle, Dec 17, 2019
2. ### TheOAP

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Hi Alan,
I'm Pete Hepple. 1st career was electronics (Radar & comms - RAF). Then programming (ICL, Plessey etc.).
Retired (age 84) but still program, mostly Math Subjects.
Would you like to communicate with me? eMail or this forum ?
One particular problem I have is this :
Written program to find factors. It works ok. I use all the usual techniques, including Synthetic Division, but have added my own "Deep Scan" : 2 of the given equation 'fed' by 'x' differing by 1/1000000 and then checking the resulting f(x1) and f(x2) results. No problem except where to start and end the scan ?
At the moment : 9x7 + 3x5 - 2x3 + 12x - 13 (say) would need the search to be +/-13., that is the largest of the leading coefficient or the constant
You can imagine the time taken if the leading Coefficient or constant were (say) 179 !
Any ideas ?
Pete Hepple

Last edited by a moderator: Jan 22, 2020
TheOAP, Jan 20, 2020
3. ### alan whittle

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Dec 17, 2019
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Hello Pete, First I am sorry to be so sluggish in replying, but I have been preoccupied with medical matters related my own health, which is not so rosy at the moment.
I am not at all expert on your problem, but have given it some thought, so I will just float the ideas that occur to me.
First, there are polynomials with no real solutions, such as x^2 +2, but I will leave them alone for now, and talk about those that do, such as all polynomials of odd degree.
If you find two values of x, say x1,x2, that give values of f(x) with opposite sign, then let x3=(x1+x2)/2, , then calculate f(x3), this must be either zero or have the same sign as either f(x1) or f(x2), so you can make a new pair x3 with x1 or x2 to repeat the cycle, and so on. This converges towards a solution to f(x)=0. If you can't find the suitable x1, x2 to get started I'm not sure what to do, but I hope this helps.
My e-mail is [email protected], so use that if you like, but next week I'm going to be in hospital some of the time. Won't have time for much maths. However, I'll be in touch when I can. I'd like to show a few of my own ideas at some time.
As a matter of interest, where do you live? I am in Barnstaple, in North Devon. I see we're the same age.

alan whittle, Jan 24, 2020
4. ### TheOAP

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Hi Alan,
I was born in Portsmouth, went to live in Germany for 4 years (Father in army), did apprenticeship in RAF. In civvy street was briefly a computer hardware tech then a OS support (ICL 2900 series).
Became a programmer, retired but still program mainly Math subjects.
Live in Mudeford, near Christchurch Dorset. I'm 84 in Feb (Leap year day...)
Hope things go well for you in hospital. You, like me seem to be a frequent visitor!

TheOAP, Jan 25, 2020
5. ### TheOAP

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Hi Alan,
I hope you're out of hospital now...
Just completed 6 visits to medical types (ear ache) and not one of them mentioned the problem !!
Pete

Last edited: Feb 27, 2020
TheOAP, Feb 27, 2020

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