Can someone help me write this into a series?
4 ビュー (過去 30 日間)
古いコメントを表示
S=2*r^3 + 4*r^5 + 6*r^7+ 8*r^9 + ... + (2n)*r^(2n+1)
0 件のコメント
回答 (3 件)
Stephen23
2018 年 11 月 3 日
編集済み: Stephen23
2018 年 11 月 3 日
r = pi;
n = 4;
S = sum((2:2:2*n).*r.^(3:2:2*n+1))
Or
r = pi;
n = 4;
v = 2:2:2*n;
sum(v.*r.^(v+1))
5 件のコメント
Stephen23
2020 年 5 月 8 日
I don't see why you need any loop:
>> n = 5;
>> (n*(n+1))/2
ans = 15
Zackary Fillbrandt
2020 年 5 月 9 日
thank you Steven, I'm not sure if what i did to acamplish my is what you ment, but just you saying that clicked on the light buld and i was able to salve the exercise. this is what i put:
function total = mysum2 (n)
total = 0
i = 1;
while i <= n
i = (n + 1);
total = n * i;
totalt = total / 2;
end
end
whith this i was able to create my 'mysum2.m' function and complete the exercise. so that you so much Steven
KALYAN ACHARJYA
2018 年 11 月 3 日
編集済み: KALYAN ACHARJYA
2018 年 11 月 3 日
syms r
n=input('Enter the n number ');
s=0;
for n=1:n;
s=2*n*r^(2*n+1)+s;
end
disp(s);
1 件のコメント
Zackary Fillbrandt
2020 年 5 月 8 日
Im having problems with an assiengment i was given for a Into to BME course:
i have to add this equation "(n(n+1))/2" into matlab without a while loop. so that my saved file "mysum2(n)" represents;
>> mysum2(5)
ans =
15
madhan ravi
2018 年 11 月 3 日
You can tell whether the series diverges or converges by the below
syms r n
SERIES=symsum((2*n)*r^(2*n+1),n,1,inf);
fplot(SERIES)
2 件のコメント
madhan ravi
2018 年 11 月 3 日
Or you can try the following if you want to see numbers as the sum:
syms r
rvalue = input('r value ? ')
n1 = input('nth term ? ')
n=1:n1;
SERIES=sum((2.*n).*r.^(2.*n+1)) %to see series in symbolic form like your question
SERIES=vpa(subs(SERIES,r,rvalue),3) %to see number with three decimal places
Zackary Fillbrandt
2020 年 5 月 8 日
Im having problems with an assiengment i was given for a Into to BME course:
i have to add this equation "(n(n+1))/2" into matlab without a while loop. so that my saved file "mysum2(n)" represents;
>> mysum2(5)
ans =
15
参考
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!